We can begin studying thermal properties of matter with definitions of temperature and heat. Temperature is a relative measure, or indication of hotness or coldness. A hot utensil is said to have a high temperature, and ice cube to have a low temperature. An object that has a higher temperature than another object is said to be hotter. Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch. However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes.

 

We know from experience that a glass of ice-cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down. It means that when the temperature of body, ice-cold water or hot tea in this case, and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium, until the body and the surrounding medium are at the same temperature. We also know that in the case of glass tumbler of ice-cold water, heat flows from the environment to the glass tumbler, whereas in the case of hot tea, it flows from the cup of hot tea to the environment. So, we can say that heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is expressed in joule (J) while SI unit of temperature is Kelvin (K), and degree Celsius (^oC) is a commonly used unit of temperature. When an object is heated, many changes may take place. Its temperature may rise, it may expand or change state. We will study the effect of heat on different bodies in later sections.

 

MEASUREMENT OF TEMPERATURE

A measure of temperature is obtained using a thermometer. Many physical properties of materials change sufficiently with temperature. Some such properties are used as the basis for constructing thermometers. The commonly used property is variation of the volume of a liquid with temperature. For example, in common liquid–in–glass thermometers, mercury, alcohol etc., are used whose volume varies linearly with temperature over a wide range.

Thermometers are calibrated so that a numerical value may be assigned to a given temperature in an appropriate scale. For the definition of any standard scale, two fixed reference points are needed. Since all substances change dimensions with temperature, an absolute reference for expansion is not available. However, the necessary fixed points may be correlated to the physical phenomena that always occur at the same temperature. The ice point and the steam point of water are two convenient fixed points and are known as the freezing and boiling points, respectively. These two points are the temperatures at which pure water freezes and boils under standard pressure. The two familiar temperature scales are the Fahrenheit temperature scale and the Celsius temperature scale. The ice and steam point have values 32 °F and 212 °F, respectively, on the Fahrenheit scale and 0 °C and 100 °C on the Celsius scale. On the Fahrenheit scale, there are 180 equal intervals between two reference points, and on the Celsius scale, there are 100.

 

A relationship for converting between the two scales may be obtained from a graph of Fahrenheit temperature (t_F) versus celsius temperature (t_C) in a straight line (Fig. 11.1), whose equation is 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG0bWaaSbaaSqaaiaadAgaaeqaaOGaeyOeI0IaaG4maiaaikdaaeaa % caaIXaGaaGioaiaaicdaaaGaeyypa0ZaaSaaaeaacaWG0bWaaSbaaS % qaaiaadogaaeqaaaGcbaGaaGymaiaaicdacaaIWaaaaaaa!4219! \frac{{{t_f} - 32}}{{180}} = \frac{{{t_c}}}{{100}}\)                   (11.1)

 

 

Fig. 11.1 A plot of Fahrenheit temperature (t_F) versus Celsius temperature (t_c).

Table 11.2 gives the values of coefficient of volume expansion of some common substances in the temperature range 0–100 °C. You can see that thermal expansion of these substances (solids and liquids) is rather small, with material, like pyrex glass and invar (a special iron-nickel alloy) having particularly low values of \(\alpha\)_V. From this Table we find that the value of \(\alpha\)_v for alcohol (ethanol) is more than mercury and expands more than mercury for the same rise in temperature.

 

 

Table 11.2  Values of coefficient of volume expansion for some substances

 

Water exhibits an anomalous behaviour; it contracts on heating between 0 °C and 4 °C. The volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4 °C, [Fig. 11.7(a)]. Below 4 °C, the volume increases, and therefore, the density decreases [Fig. 11.7(b)].

 

This means that water has the maximum density at 4 °C. This property has an important environmental effect: bodies of water, such as 

 

Fig. 11.7 Thermal expansion of water

 

lakes and ponds, freeze at the top first. As a lake cools toward 4 °C, water near the surface loses energy to the atmosphere, becomes denser, and sinks; the warmer, less dense water near the bottom rises. However, once the colder water on top reaches temperature below 4 °C, it becomes less dense and remains at the surface, where it freezes. If water did not have this property, lakes and ponds would freeze from the bottom up, which would destroy much of their animal and plant life.

 

Gases, at ordinary temperature, expand more than solids and liquids. For liquids, the coefficient of volume expansion is relatively independent of the temperature. However, for gases it is dependent on temperature. For an ideal gas, the coefficient of volume expansion at constant pressure can be found from the ideal gas equation:

PV=\(\mu\)RT

At constant pressure

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGqb % GaeyiLdqKaamOvaiabg2da9iabeY7aTjaadkfacqGHuoarcaWGubaa % baWaaSaaaeaacqGHuoarcaWGwbaabaGaamOvaaaacqGH9aqpdaWcaa % qaaiabgs5aejaadsfaaeaacaWGubaaaaqaaiaadMgacaGGUaGaamyz % aiaacUdacqaHXoqydaWgaaWcbaGaamODaaqabaGccqGH9aqpdaWcaa % qaaiaaigdaaeaacaWGubaaaiaaykW7caaMc8UaaGPaVlaadAgacaWG % VbGaamOCaiaaykW7caaMc8UaaGPaVlaadMgacaWGKbGaamyzaiaadg % gacaWGSbGaaGPaVlaaykW7caaMc8Uaam4zaiaadggacaWGZbGaaGPa % VlaaykW7caaMc8+aaeWaaeaacaaIXaGaaGymaiaac6cacaaI2aaaca % GLOaGaayzkaaaaaaa!7034! \begin{array}{l} P\Delta V = \mu R\Delta T\\ \frac{{\Delta V}}{V} = \frac{{\Delta T}}{T}\\ i.e;{\alpha _v} = \frac{1}{T}\,\,\,for\,\,\,ideal\,\,\,gas\,\,\,\left( {11.6} \right) \end{array}\)

 

At 0 °C, \(\alpha\)_v = 3.7 × 10^–3 K^–1, which is much larger than that for solids and liquids. Equation (11.6) shows the temperature dependence of \(\alpha\)_v; it decreases with increasing temperature. For a gas at room temperature and constant pressure, \(\alpha\)_v is about 3300 × 10^–6 K^–1, as much as order(s) of magnitude larger than the coefficient of volume expansion of typical liquids. There is a simple relation between the coefficient of volume expansion (\(\alpha\)_v) and coefficient of linear expansion (\(\alpha\)₁). Imagine a cube of length, l, that expands equally in all directions, when its temperature increases by \(\Delta\)T. We have 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuo % arcaWGSbGaeyypa0JaeqySde2aaSbaaSqaaiaaigdaaeqaaOGaamiB % aiabgs5aejaadsfaaeaacaWGZbGaam4BaiaacYcacaaMc8UaaGPaVl % abgs5aejaadAfacqGH9aqpdaqadaqaaiaadYgacqGHRaWkcqGHuoar % caWGSbaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0 % IaamiBamaaCaaaleqabaGaaG4maaaakiabloKi7iaaiodacaWGSbWa % aWbaaSqabeaacaaIYaaaaOGaeyiLdqKaamiBaiaaykW7caaMc8UaaG % PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM % c8UaaGPaVlaaykW7daqadaqaaiaaigdacaaIXaGaaiOlaiaaiEdaai % aawIcacaGLPaaaaaaa!703B! \begin{array}{l} \Delta l = {\alpha _1}l\Delta T\\ so,\,\,\Delta V = {\left( {l + \Delta l} \right)^3} - {l^3} \simeq 3{l^2}\Delta l\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {11.7} \right) \end{array}\)

In Equation (11.7), terms in (\(\Delta\)l)² and (\(\Delta\)l)³ have been neglected since \(\Delta\)l is small compared to l. So

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % Ovaiabg2da9maalaaabaGaaG4maiaadAfacqGHuoarcaWGSbaabaGa % amiBaaaacqGH9aqpcaaIZaGaamOvaiabeg7aHnaaBaaaleaacaWGSb % aabeaakiabgs5aejaadsfaaaa!45D3! \Delta V = \frac{{3V\Delta l}}{l} = 3V{\alpha _l}\Delta T\)              (11.18)

which gives

 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySde2aaS % baaSqaaiaadAhaaeqaaOGaeyypa0JaaG4maiabeg7aHnaaBaaaleaa % caaIXaaabeaaaaa!3D0F! {\alpha _v} = 3{\alpha _1}\)                                     (11.19)

What happens by preventing the thermal expansion of a rod by fixing its ends rigidly? Clearly, the rod acquires a compressive strain due to the external forces provided by the rigid support at the ends. The corresponding stress set up in the rod is called thermal stress. For example, consider a steel rail of length 5 m and area of cross-section 40 cm² that is prevented from expanding while the temperature rises by 10 °C. The coefficient of linear expansion of steel is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySde2aaS % baaSqaamaabmaabaGaam4CaiaadshacaWGLbGaamyzaiaadYgaaiaa % wIcacaGLPaaaaeqaaOGaeyypa0JaaGymaiaac6cacaaIYaGaey41aq % RaaiiyaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaGccaWGlbWa % aWbaaSqabeaacqGHsislcaaIXaaaaaaa!4976! {\alpha _{\left( {steel} \right)}} = 1.2 \times `{0^{ - 5}}{K^{ - 1}}\). Thus, the compressive strain is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHuoarcaWGSbaabaGaamiBaaaacqGH9aqpcqaHXoqydaWgaaWcbaGa % aGymamaabmaabaGaam4CaiaadshacaWGLbGaamyzaiaadYgaaiaawI % cacaGLPaaaaeqaaOGaeyiLdqKaamivaiabg2da9iaaigdacaGGUaGa % aGOmaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI1a % aaaOGaey41aqRaaGymaiaaicdacqGH9aqpcaaIXaGaaiOlaiaaikda % cqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGinaaaaaa % a!5A21! \frac{{\Delta l}}{l} = {\alpha _{1\left( {steel} \right)}}\Delta T = 1.2 \times {10^{ - 5}} \times 10 = 1.2 \times {10^{ - 4}}\). 

Youngs modulus of steel is Y_(steel)= 2 × 10¹¹ N m^–2. Therefore, the thermal stress developed is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHuoarcaWGgbaabaGaamyqaaaacqGH9aqpcaWGubWaaSbaaSqaaiaa % dohacaWG0bGaamyzaiaadwgacaWGSbaabeaakmaabmaabaWaaSaaae % aacqGHuoarcaWGSbaabaGaamiBaaaaaiaawIcacaGLPaaacqGH9aqp % caaIYaGaaiOlaiaaisdacqGHxdaTcaaIXaGaaGimamaaCaaaleqaba % GaaG4naaaakiaad6eacaWGTbWaaWbaaSqabeaacqGHsislcaaIYaGa % aiilaaaaaaa!50AC! \frac{{\Delta F}}{A} = {T_{steel}}\left( {\frac{{\Delta l}}{l}} \right) = 2.4 \times {10^7}N{m^{ - 2,}}\) which, corresponds to an external force of \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % Oraiabg2da9iaadgeacaWGzbWaaSbaaSqaaiaadohacaWG0bGaamyz % aiaadwgacaWGSbaabeaakmaabmaabaWaaSaaaeaacqGHuoarcaWGSb % aabaGaamiBaaaaaiaawIcacaGLPaaacqGH9aqpcaaIYaGaaiOlaiaa % isdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaG4naaaakiabgE % na0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaOGaeS4q % ISJaaGymaiaaicdadaahaaWcbeqaaiaaiwdaaaGccaWGobaaaa!5633! \Delta F = A{Y_{steel}}\left( {\frac{{\Delta l}}{l}} \right) = 2.4 \times {10^7} \times {10^{ - 4}} \simeq {10^5}N\).

If two such steel rails, fixed at their outer ends, are in contact at their inner ends, a force of this magnitude can easily bend the rails.

 

EXAMPLE 1

 

Show that the coefficient of area expansion, ( \(\Delta\)A/A)/ \(\Delta\)T, of a rectangular sheet of the solid is twice its linear expansivity, \(\alpha\)₁.

 

ANSWER

 

Consider a rectangular sheet of the solid material of length a and breadth b (Fig. 11.8 ). When the temperature increases by \(\Delta\)T, a increases by \(\Delta\)a = \(\alpha_1\)a\(\Delta\)T and b increases by \(\Delta\)b = \(\alpha_1\)b \(\Delta\)T. From Fig. 11.8, the increase in area

 

 

                   Fig. 11.8

 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuo % arcaWGbbGaeyypa0JaeyiLdqKaamyqamaaBaaaleaacaaIXaaabeaa % kiabgUcaRiabgs5aejaadgeadaWgaaWcbaGaaGOmaaqabaGccqGHRa % WkcqGHuoarcaWGbbWaaSbaaSqaaiaaiodaaeqaaaGcbaGaeyiLdqKa % amyqaiabg2da9iaadggacqGHuoarcaWGIbGaey4kaSIaamOyaiabgs % 5aejaadggacqGHRaWkdaqadaqaaiabgs5aejaadggaaiaawIcacaGL % Paaadaqadaqaaiabgs5aejaadkgaaiaawIcacaGLPaaaaeaacqGH9a % qpcaWGHbGaeqySde2aaSbaaSqaaiaaigdaaeqaaOGaamOyaiabgs5a % ejaadsfacqGHRaWkcaWGIbGaeqySde2aaSbaaSqaaiaaigdaaeqaaO % Gaamyyaiabgs5aejaadsfacqGHRaWkdaqadaqaaiabeg7aHnaaBaaa % leaacaaIXaaabeaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaa % aakiaadggacaWGIbWaaeWaaeaacqGHuoarcaWGubaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaaGcbaGaeyypa0JaeqySde2aaSbaaS % qaaiaaigdaaeqaaOGaamyyaiaadkgacqGHuoarcaWGubWaaeWaaeaa % caaIYaGaey4kaSIaeqySde2aaSbaaSqaaiaaigdaaeqaaOGaeyiLdq % KaamivaaGaayjkaiaawMcaaiabg2da9iabeg7aHnaaBaaaleaacaaI % XaaabeaakiaadgeacqGHuoarcaWGubWaaeWaaeaacaaIYaGaey4kaS % IaeqySde2aaSbaaSqaaiaaigdaaeqaaOGaeyiLdqKaamivaaGaayjk % aiaawMcaaaaaaa!9125! \begin{array}{l} \Delta A = \Delta {A_1} + \Delta {A_2} + \Delta {A_3}\\ \Delta A = a\Delta b + b\Delta a + \left( {\Delta a} \right)\left( {\Delta b} \right)\\ = a{\alpha _1}b\Delta T + b{\alpha _1}a\Delta T + {\left( {{\alpha _1}} \right)^2}ab{\left( {\Delta T} \right)^2}\\ = {\alpha _1}ab\Delta T\left( {2 + {\alpha _1}\Delta T} \right) = {\alpha _1}A\Delta T\left( {2 + {\alpha _1}\Delta T} \right) \end{array}\)

Since \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySde2aaS % baaSqaaiaaigdaaeqaaOGaeS4qISdaaa!39B7! {\alpha _1} \simeq \)10^–5 K^–1, from Table 11.1, the product \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySde2aaS % baaSqaaiaaigdaaeqaaOGaeyiLdqeaaa!39ED! {\alpha _1}\Delta \)T for fractional temperature is small in comparision with 2 and may be neglected.

Hence,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % Wcaaqaaiabgs5aejaadgeaaeaacaWGbbaaaaGaayjkaiaawMcaamaa % laaabaGaaGymaaqaaiabgs5aejaadsfaaaGaeS4qISJaaGOmaiabeg % 7aHnaaBaaaleaacaaIXaaabeaaaaa!4200! \left( {\frac{{\Delta A}}{A}} \right)\frac{1}{{\Delta T}} \simeq 2{\alpha _1}\)

 

EXAMPLE 2

A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m, respectively at 27 °C. To what temperature should the ring be heated so as to fit the rim of the wheel?

 

ANSWER

 

Given T₁= 27 °C

          T₂ = 5.231 m

          \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamitamaaBa % aaleaacaWGubWaaSbaaWqaaiaaikdaaeqaaaWcbeaaaaa!38C0! {L_{{T_2}}}\) = 5.243 m

So, 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGmb % WaaSbaaSqaaiaadsfadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyyp % a0JaamitamaaBaaaleaacaWGubWaaSbaaWqaaiaaigdaaeqaaaWcbe % aakiabg2da9maadmaabaGaaGymaiabgUcaRiabeg7aHnaaBaaaleaa % caaIXaaabeaakmaabmaabaGaamivamaaBaaaleaacaaIYaaabeaaki % abgkHiTiaadsfadaWgaaWcbaGaaGymaaqabaaakiaawIcacaGLPaaa % aiaawUfacaGLDbaaaeaacaaI1aGaaiOlaiaaikdacaaI0aGaaG4mai % aaykW7caWGTbGaeyypa0JaaGynaiaac6cacaaIYaGaaG4maiaaigda % caaMc8UaaGPaVlaad2gacaaMc8+aamWaaeaacaaIXaGaey4kaSIaaG % ymaiaac6cacaaIYaGaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqa % beaacqGHsislcaaI1aaaaOGaam4samaaCaaaleqabaGaeyOeI0IaaG % ymaaaakmaabmaabaGaamivamaaBaaaleaacaaIYaaabeaakiabgkHi % TiaaikdacaaI3aWaaWbaaSqabeaacqWIyiYBaaGccaWGdbaacaGLOa % GaayzkaaaacaGLBbGaayzxaaGaaGPaVlaaykW7caaMc8Uaam4Baiaa % dkhacaaMc8UaaGPaVlaadsfadaWgaaWcbaGaaGOmaaqabaGccqGH9a % qpcaaIYaGaaGymaiaaiIdadaahaaWcbeqaaiablIHiVbaakiaadoea % aaaa!81A9! \begin{array}{l} {L_{{T_2}}} = {L_{{T_1}}} = \left[ {1 + {\alpha _1}\left( {{T_2} - {T_1}} \right)} \right]\\ 5.243\,m = 5.231\,\,m\,\left[ {1 + 1.20 \times {{10}^{ - 5}}{K^{ - 1}}\left( {{T_2} - {{27}^ \circ }C} \right)} \right]\,\,\,or\,\,{T_2} = {218^ \circ }C \end{array}\)

 

SPECIFIC HEAT CAPACITY

Take some water in a vessel and start heating it on a burner. Soon you will notice that bubbles begin to move upward. As the temperature is raised the motion of water particles increases till it becomes turbulent as water starts boiling. What are the factors on which the quantity of heat required to raise the temperature of a substance depend? In order to answer this question in the first step, heat a given quantity of water to raise its temperature by, say 20 °C and note the time taken. Again take the same amount of water and raise its temperature by 40 °C using the same source of heat. Note the time taken by using a stopwatch. You will find it takes about twice the time and therefore, double the quantity of heat required raising twice the temperature of same amount of water.

 

In the second step, now suppose you take double the amount of water and heat it, using the same heating arrangement, to raise the temperature by 20 °C, you will find the time taken is again twice that required in the first step.

 

In the third step, in place of water, now heat the same quantity of some oil, say mustard oil, and raise the temperature again by 20 °C. Now 

note the time by the same stopwatch. You will find the time taken will be shorter and therefore, the quantity of heat required would be less than that required by the same amount of water for the same rise in temperature.

 

The above observations show that the quantity of heat required to warm a given substance depends on its mass, m, the change in temperature, \(\Delta\)T and the nature of substance. The change in temperature of a substance, when a given quantity of heat is absorbed or rejected by it, is characterised by a quantity called the heat capacity of that substance. We define heat capacity, S of a substance as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaeyiLdqKaamyuaaqaaiabgs5aejaadsfaaaaaaa!3C61! S = \frac{{\Delta Q}}{{\Delta T}}\)                                (11.10)

 

where \(\Delta\)Q is the amount of heat supplied to the substance to change its temperature from T to T + \(\Delta\)T.

You have observed that if equal amount of heat is added to equal masses of different substances, the resulting temperature changes will not be the same. It implies that every substance has a unique value for the amount of heat absorbed or given off to change the temperature of unit mass of it by one unit. This quantity is referred to as the specific heat capacity of the substance.

 

If \(\Delta\)Q stands for the amount of heat absorbed or given off by a substance of mass m when it undergoes a temperature change T, then the specific heat capacity, of that substance is given by 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Caiabg2 % da9maalaaabaGaam4uaaqaaiaad2gaaaGaeyypa0ZaaSaaaeaacaaI % XaaabaGaamyBaaaadaWcaaqaaiabgs5aejaadgfaaeaacqGHuoarca % WGubaaaaaa!411E! s = \frac{S}{m} = \frac{1}{m}\frac{{\Delta Q}}{{\Delta T}}\)                   (11.11)

 

The specific heat capacity is the property of the substance which determines the change in the temperature of the substance (undergoing no phase change) when a given quantity of heat is absorbed (or given off) by it. It is defined as the amount of heat per unit mass absorbed or given off by the substance to change its temperature by one unit. It depends on the nature of the substance and its temperature. The SI unit of specific heat capacity is J kg^–1 K^–1.

 

If the amount of substance is specified in terms of moles \(\mu\), instead of mass m in kg, we can define heat capacity per mole of the substance by

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiabg2 % da9maalaaabaGaam4uaaqaaiabeY7aTbaacqGH9aqpdaWcaaqaaiaa % igdaaeaacqaH8oqBaaWaaSaaaeaacqGHuoarcaWGrbaabaGaeyiLdq % Kaamivaaaaaaa!4276! C = \frac{S}{\mu } = \frac{1}{\mu }\frac{{\Delta Q}}{{\Delta T}}\)                                (11.12)

where C is known as molar specific heat capacity of the substance. Like S, C also depends on the nature of the substance and its temperature. The SI unit of molar specific heat capacity is J mol^–1 K^–1.

 

However, in connection with specific heat capacity of gases, additional conditions may be needed to define C. In this case, heat transfer can be achieved by keeping either pressure or volume constant. If the gas is held under constant pressure during the heat transfer, then it is called the molar specific heat capacity at constant pressure and is denoted by C_p. On the other hand, if the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume and is denoted by C_v. For details see Chapter 12. Table 11.3 lists measured specific heat capacity of some substances at atmospheric pressure and ordinary temperature while Table 11.4 lists molar specific heat capacities of some gases. From Table 11.3 you can note that water 

 

Table 11.3  Specific heat capacity of some substances at room temperature and atmospheric pressure

 

 

has the highest specific heat capacity compared to other substances. For this reason water is also used as a coolant in automobile radiators, as well as, a heater in hot water bags. Owing to its high specific heat capacity, water warms up more slowly than land during summer, and consequently wind from the sea has a cooling effect. Now, you can tell why in desert areas, the earth surface warms up quickly during the day and cools quickly at night.

 

Table 11.4 Molar specific heat capacities of some gases

 

 

CALORIMETRY

 

A system is said to be isolated if no exchange or transfer of heat occurs between the system and its surroundings. When different parts of an isolated system are at different temperature, a quantity of heat transfers from the part at higher temperature to the part at lower temperature. The heat lost by the part at higher temperature is equal to the heat gained by the part at lower temperature.

 

Calorimetry means measurement of heat. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, provided no heat is allowed to escape to the surroundings. A device in which heat measurement can be done is called a calorimeter. It consists of a metallic vessel and stirrer of the same material, like copper or aluminium. The vessel is kept inside a wooden jacket, which contains heat insulating material, like glass wool etc. The outer jacket acts as a heat shield and reduces the heat loss from the inner vessel. There is an opening in the outer jacket through which a mercury thermometer can be inserted into the calorimeter (Fig. 11.20).

The following example provides a method by which the specific heat capacity of a given solid can be determinated by using the principle, heat gained is equal to the heat lost.

 

EXAMPLE 3

 

A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.

 

ANSWER

 

In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.

Mass of aluminium sphere (m₁) = 0.047 kg Initial temperature of aluminium sphere=100 °C Final temperature = 23 °C

Change in temperature (\(\Delta\)T)=(100 °C - 23 °C) = 77 °C

Let specific heat capacity of aluminium be s_Al.

The amount of heat lost by the aluminium sphere = m₁s_Al \(\Delta\)T = 0.047kg \(\times\) s_Al \(\times\) 77\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeSigI8gaaa!3730! \circ \) C

Mass of water (m₂) = 0.25 kg

Mass of calorimeter (m₃) = 0.14 kg

Initial temperature of water and calorimeter=20 °C Final temperature of the mixture = 23 °C Change in temperature (\(\Delta\)T₂) = 23 °C – 20 °C = 3 °C

Specific heat capacity of water (s_w)

= 4.18 × 10³ J kg^–1 K^–1

Specific heat capacity of copper calorimeter

= 0.386 × 10³ J kg^–1 K^–1

The amount of heat gained by water and calorimeter = m₂ s_w \(\Delta\)T₂ + m₃s_cu\(\Delta\)T₂

= (m₂s_w + m₃s_cu) (\(\Delta\)T₂)

= (0.25 kg × 4.18 × 10³ J kg^–1 K^–1 + 0.14 kg × 0.386 × 10³ J kg^–1 K^–1) (23 °C – 20 °C)

In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.

So, 0.047 kg × s_Al × 77 °C

= (0.25 kg × 4.18 × 10³ J kg^–1 K^–1+ 0.14 kg × 0.386 × 10³ J kg^–1 K^–1)(3 °C)

s_Al = 0.911 kJ kg ^–1 K^–1

 

 

CHANGE OF STATE

Matter normally exists in three states: solid, liquid and gas. A transition from one of these states to another is called a change of state. Two common changes of states are solid to liquid and liquid to gas (and, vice versa). These changes can occur when the exchange of heat takes place between the substance and its surroundings. To study the change of state on heating or cooling, let us perform the following activity.

 

Take some cubes of ice in a beaker. Note the temperature of ice. Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Draw a graph between temperature and time (Fig. 11.9).

 

 

Fig. 11.9 A plot of temperature versus time showing the changes in the state of ice on heating (not to scale).

 

You will observe no change in the temperature as long as there is ice in the beaker. In the above process, the temperature of the system does not change even though heat is being continuously supplied. The heat supplied is being utilised in changing the state from solid (ice) to liquid (water).

 

The change of state from solid to liquid is called melting and from liquid to solid is called fusion. It is observed that the temperature remains constant until the entire amount of the solid substance melts. That is, both the solid and the liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid. The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each other is called its melting point. It is characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atomspheric pressure is called its normal melting point. Let us do the following activity to understand the process of melting of ice.

 

Take a slab of ice. Take a metallic wire and fix two blocks, say 5 kg each, at its ends. Put the wire over the slab as shown in Fig. 11.10.

 

                      Fig. 11.10.

 

You will observe that the wire passes through the ice slab. This happens due to the fact that just below the wire, ice melts at lower temperature due to increase in pressure. When the wire has passed, water above the wire freezes again. Thus, the wire passes through the slab and the slab does not split. This phenomenon of refreezing is called regelation. Skating is possible on snow due to the formation of water under the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

 

After the whole of ice gets converted into water and as we continue further heating, we shall see that temperature begins to rise (Fig.11.9). The temperature keeps on rising till it reaches nearly 100 °C when it again becomes steady. The heat supplied is now being utilised to change water from liquid state to vapour or gaseous state.

 

The change of state from liquid to vapour (or gas) is called vaporisation. It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour. That is, both the liquid and vapour states of the substance coexist in thermal equilibrium, during the change of state from liquid to vapour. The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point. Let us do the following activity to understand the process of boiling of water.

 

Take a round-bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask (Fig. 11.11).

 

Fig. 11.11 Boiling process

 

If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling begins again. Thus boiling point increases with increase in pressure.

Let us now remove the burner. Allow water to cool to about 80 °C. Remove the thermometer and steam outlet. Close the flask with the airtight 

cork. Keep the f lask turned upside down on the stand. Pour ice-cold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure.

This explains why cooking is difficult on hills. At high altitudes, atmospheric pressure is lower, reducing the boiling point of water as compared to that at sea level. On the other hand, boiling point is increased inside a pressure cooker by increasing the pressure. Hence cooking is faster. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point.

 

However, all substances do not pass through the three states: solid-liquid-gas. There are certain substances which normally pass from the solid to the vapour state directly and vice versa. The change from solid state to vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime. Dry ice (solid CO₂) sublimes, so also iodine. During the sublimation process both the solid and vapour states of a substance coexist in thermal equilibrium.

 

Latent Heat

 

In Section 11.8, we have learnt that certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state. The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process. For example, if heat is added to a given quantity of ice at –10 °C, the temperature of ice increases until it reaches its melting point (0 °C). At this temperature, the addition of more heat does not increase the temperature but causes the ice to melt, or changes its state. Once the entire ice melts, adding more heat will cause the temperature of the water to rise. A similar situation occurs during liquid gas change of state at the boiling point. Adding more heat to boiling water causes vaporisation, without increase in temperature .

 

The heat required during a change of state depends upon the heat of transformation and the mass of the substance undergoing a change of state. Thus, if mass m of a substance undergoes a change from one state to the other, then the quantity of heat required is given by

Q = m L

or   L = Q/m                                       (11.13)

where L is known as latent heat and is a characteristic of the substance. Its SI unit is J kg^–1. The value of L also depends on the pressure. Its value is usually quoted at standard atmospheric pressure. The latent heat for a solid- liquid state change is called the latent heat of 

fusion (L_f), and that for a liquid-gas state change is called the latent heat of vaporisation (L_v). These are often referred to as the heat of fusion and the heat of vaporisation. A plot of temperature versus heat for a quantity of water is shown in Fig. 11.12.

 

Fig. 11.12 Temperature versus heat for water at 1 atm pressure (not to scale).

 

The latent heats of some substances, their freezing and boiling points, are given in Table 11.5.

 

Table 11.5   Temperatures of the change of state and latent heats for various substances at 1 atm pressure

 

Note that when heat is added (or removed) during a change of state, the temperature remains constant. Note in Fig. 11.12 that the slopes of the phase lines are not all the same, which indicate that specific heats of the various states are not equal. For water, the latent heat of fusion and vaporisation are L_f = 3.33 × 10⁵ J kg^–1 and L_v = 22.6 × 10⁵ J kg^–1, respectively. That is, 3.33 × 10⁵ J of heat is needed to melt 1 kg ice at0 °C, and 22.6 × 10⁵ J of heat is needed to convert 1 kg water into steam at 100 °C. So, steam at 100 °C carries 22.6 × 10⁵ J kg^–1 more heat than water at 100 °C. This is why burns from steam are usually more serious than those from boiling water.

 

EXAMPLE 4

 

When 0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Calculate the heat of fusion of ice. (s_water = 4186 J kg^–1 K^–1)

 

ANSWER

 

Heat lost by water = msw (\(\theta\)_f–\(\theta\)_i)w

= (0.30 kg) (4186 J kg^–1 K^–1) (50.0 °C – 6.7 °C)

= 54376.14 J

Heat required to melt ice = m₂L_f = (0.15 kg) L_f

Heat required to raise temperature of ice water to final temperature = m_Is_w (\(\theta\)_f–\(\theta\)_i)_I

= (0.15 kg) (4186 J kg^–1 K ^–1) (6.7 °C – 0 °C)

= 4206.93 J

Heat lost = heat gained

54376.14 J = (0.15 kg ) L_f + 4206.93 J

L_f = 3.34×10⁵ J kg^–1.

 

EXAMPLE 5

 

Calculate the heat required to convert 3 kg of ice at –12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^–1 K^–1, specific heat capacity of water = 4186 J kg^{– 1} K^–1, latent heat of fusion of ice = 3.35 × 10⁵ J kg^–1 and latent heat of steam = 2.256 ×10⁶ J kg^–1.

 

ANSWER

 

We have

Mass of the ice, m = 3 kg specific heat capacity of ice, s_ice

= 2100 J kg^–1 K^–1

specific heat capacity of water, s_water

= 4186 J kg^–1 K^–1

latent heat of fusion of ice, L_{f ice}

= 3.35 × 10⁵ J kg^–1

latent heat of steam, L_steam

= 2.256 × 10⁶ J kg^–1

Now, Q = heat required to convert 3 kg of ice at –12 °C to steam at 100 °C,

Q₁  = heat required to convert ice at

–12 °C to ice at 0 °C.

= m sice \(\Delta\)T₁ = (3 kg) (2100 J kg–1.K^–1) [0–(–12)]°C = 75600 J

Q₂  = heat required to melt ice at 0 °C to water at 0 °C

= m L_{f ice} = (3 kg) (3.35 × 10⁵ J kg^–1)

=    1005000 J

Q₃  = heat required to convert water at 0 °C to water at 100 °C.

= msw \(\Delta\)T₂ = (3kg) (4186J kg^–1 K^–1) (100 °C)

= 1255800 J

Q₄  = heat required to convert water at 100 °C to steam at 100 °C

= m L_steam = (3 kg) (2.256×10⁶J kg^–1)

=   6768000 J

So, Q  = Q₁ + Q₂ + Q₃ + Q₄

= 75600J + 1005000 J+ 1255800 J + 6768000 J

= 9.1×106 J

 

HEAT TRANSFER

We have seen that heat is energy transfer from one system to another or from one part of a system to another part, arising due to temperature difference. What are the different ways by which this energy transfer takes place? There are three distinct modes of heat transfer: conduction, convection and radiation (Fig. 11.13).

 

 

Fig. 11.13  Heating by conduction, convection and radiation

 

Conduction

Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference. Suppose, one end of a metallic rod is put in a flame, the other end of the rod will soon be so hot that you cannot hold it by your bare hands. Here, heat transfer takes place by conduction from the hot end of the rod through its different parts to the other end. Gases are poor thermal conductors, while liquids have conductivities intermediate between solids and gases.

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference. Consider a metallic bar of length L and uniform cross-section A with its two ends maintained at different temperatures. This can be done, for example, by putting the ends in thermal contact with large reservoirs at temperatures, say, T_C and T_D, respectively (Fig. 11.14). 

Fig. 11.14 Steady state heat flow by conduction in a bar with its two ends maintained at temperatures T_C and T_D; (T_C > T_D).

Let us assume the ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.

After sometime, a steady state is reached; the temperature of the bar decreases uniformly with distance from T_C to T_D; (T_C>T_D). The reservoir at

C supplies heat at a constant rate, which transfers through the bar and is given out at the same rate to the reservoir at D. It is found experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional to the temperature difference (T_C – T_D) and the area of cross-section A and is inversely proportional to the length L :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiabg2 % da9iaadUeacaWGbbWaaSaaaeaacaWGubWaaSbaaSqaaiaadoeaaeqa % aOGaeyOeI0IaamivamaaBaaaleaacaWGebaabeaaaOqaaiaadYeaaa % aaaa!3EDC! H = KA\frac{{{T_C} - {T_D}}}{L}\)           (11.14)

 

The constant of proportionality K is called the thermal conductivity of the material. The greater the value of K for a material, the more rapidly will it conduct heat. The SI unit of K is J s^–1 m^–1 K^–1 or W m^–1 K^–1. The thermal conductivities of various substances are listed in Table 11.6.

 

Table 11.6 Thermal conductivities of some material

 

These values vary slightly with temperature, but can be considered to be constant over a normal temperature range.

 

Compare the relatively large thermal conductivities of good thermal conductors and, metals, with the relatively small thermal conductivities of some good thermal insulators, such as wood and glass wool. You may have noticed that some cooking pots have copper coating on the bottom. Being a good conductor of heat, copper promotes the distribution of heat over the bottom of a pot for uniform cooking. Plastic foams, on the other hand, are good insulators, mainly because they contain pockets of air. Recall that gases are poor conductors, and note the low thermal conductivity of air in the Table 11.5. Heat retention and transfer are important in many other applications. Houses made of concrete roofs get very hot during summer days because thermal conductivity of concrete (though much smaller than that of a metal) is still not small enough. Therefore, people, usually, prefer to give a layer of earth or foam insulation on the ceiling so that heat transfer is prohibited and keeps the room cooler. In some situations, heat transfer is critical. In a nuclear reactor, for example, elaborate heat transfer systems need to be installed so that the enormous energy produced by nuclear fission in the core transits out sufficiently fast, thus preventing the core from overheating.

 

EXAMPLE 6

 

What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 11.15. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300 °C, temperature of the other end = 0 °C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s^-1 m^-1 K^-1; and of copper = 385 J s^-1 m^-1 K^-1).

       fig. 11. 15.

ANSWER

The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise 

there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steel-copper rod. Let T be the temperature of the steel-copper junction in the steady state. Then, 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGlbWaaSbaaSqaaiaaigdaaeqaaOGaamyqamaaBaaaleaacaaIXaaa % beaakmaabmaabaGaaG4maiaaicdacaaIWaGaeyOeI0IaamivaaGaay % jkaiaawMcaaaqaaiaadYeadaWgaaWcbaGaaGymaaqabaaaaOGaeyyp % a0ZaaSaaaeaacaWGlbWaaSbaaSqaaiaaikdaaeqaaOGaamyqamaaBa % aaleaacaaIYaaabeaakmaabmaabaGaamivaiabgkHiTiaaicdaaiaa % wIcacaGLPaaaaeaacaWGmbWaaSbaaSqaaiaaikdaaeqaaaaaaaa!4B12! \frac{{{K_1}{A_1}\left( {300 - T} \right)}}{{{L_1}}} = \frac{{{K_2}{A_2}\left( {T - 0} \right)}}{{{L_2}}}\)

where 1 and 2 refer to the steel and copper rod respectively. For A₁ = 2 A₂, L₁ = 15.0 cm, L₂ = 10.0 cm, K₁ = 50.2 J s^–1 m^–1 K ^–1, K₂ = 385 J s^–1 m^–1 K ^–1, we have 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiaaiwdacaaIWaGaaiOlaiaaikdacqGHxdaTcaaIYaWaaeWaaeaa % caaIZaGaaGimaiaaicdacqGHsislcaWGubaacaGLOaGaayzkaaaaba % GaaGymaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaIZaGaaGioaiaaiwda % caWGubaabaGaaGymaiaaicdaaaaabaGaam4DaiaadIgacaWGPbGaam % 4yaiaadIgacaaMc8UaaGPaVlaaykW7caWGNbGaamyAaiaadAhacaWG % LbGaam4CaiaaykW7caaMc8UaaGPaVlaadsfacqGH9aqpcaaI0aGaaG % inaiaac6cacaaI0aWaaWbaaSqabeaacqWIyiYBaaGccaWGdbaaaaa!620C! \begin{array}{l} \frac{{50.2 \times 2\left( {300 - T} \right)}}{{15}} = \frac{{385T}}{{10}}\\ which\,\,\,gives\,\,\,T = {44.4^ \circ }C \end{array}\)

EXAMPLE 7

An iron bar (L₁ = 0.1 m, A₁ = 0.02 m₂, K₁ = 79 W m^–1 K^–1) and a brass bar (L₂ = 0.1 m, A₂ = 0.02 m², K₂ = 10⁹ W m^–1K^–1) are soldered end to end as shown in Fig. 11.16. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (iii) the heat current through the compound bar.

 

                         Fig. 11.16

 

ANSWER

 

Given, L₁ = L₂= L = 0.1 m, A₁ = A₂= A= 0.02 m² K₁ = 79 W m^–1 K^–1, K₂ = 10⁹ W m^–1 K^–1 , T₁ = 373 K, and T₂ = 273 K.

Under steady state condition, the heat current (H₁) through iron bar is equal to the heat current (H2) through brass bar.

So, H = H₁ = H₂

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGlbWaaSbaaSqaaiaaigdaaeqaaOGaamyqamaaBaaaleaa % caaIXaaabeaakmaabmaabaGaamivamaaBaaaleaacaaIXaaabeaaki % abgkHiTiaadsfadaWgaaWcbaGaaGimaaqabaaakiaawIcacaGLPaaa % aeaacaWGmbWaaSbaaSqaaiaaigdaaeqaaaaakiabg2da9maalaaaba % Gaam4samaaBaaaleaacaaIYaaabeaakiaadgeadaWgaaWcbaGaaGOm % aaqabaGcdaqadaqaaiaadsfadaWgaaWcbaGaaGimaaqabaGccqGHsi % slcaWGubWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaaabaGa = \frac{{{K_1}{A_1}\left( {{T_1} - {T_0}} \right)}}{{{L_1}}} = \frac{{{K_2}{A_2}\left( {{T_0} - {T_2}} \right)}}{{{L_2}}}\)