THERMAL PROPERTIES OF MATTER-Exercise

Q1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Solution

Given triple point temperatures: neon \(T = 24.57~\text{K}\), carbon dioxide \(T = 216.55~\text{K}\).

For neon:

For carbon dioxide:

Thus, neon triple point is \(-248.58^\circ\text{C}\) and \(-415.44^\circ\text{F}\); carbon dioxide triple point is \(-56.60^\circ\text{C}\) and \(-69.88^\circ\text{F}\).

Q2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between \(\mathrm{T_A}\) and \(\mathrm{T_B}\)?

Solution

Given: Triple point of water is \(200~\text{A}\) on scale A and \(350~\text{B}\) on scale B. Since both represent identical physical temperature \(273.15~\text{K}\), the scales relate through this common reference.

The relation between temperature readings follows the standard absolute scale form \(\frac{T}{T_{\text{tp}}} = \frac{T'}{T'_{\text{tp}}}\), where subscript tp denotes triple point value.

Thus, temperatures on scale A equal \(\frac{4}{7}\) times temperatures on scale B. For verification, ice point (273.15 K) yields \(T_A = 200 \times \frac{273.15}{273.15} = 200~\text{A}\), \(T_B = 350~\text{B}\), satisfying \(\frac{200}{350} = \frac{4}{7}\).

This linear proportion holds since both scales share identical zero points and employ triple point as fixed reference, differing only in numerical assignment to that universal temperature fixed point.

Q3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?

Solution

Q4. Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15 Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

Solution

(a) The triple-point of water is used as a standard fixed point in modern thermometry because it represents a unique and reproducible state of equilibrium between ice, liquid water, and water vapour. This equilibrium occurs at a definite temperature that is independent of atmospheric pressure and other external conditions. In contrast, the melting point of ice and the boiling point of water depend strongly on pressure and impurities. Variations in atmospheric pressure cause noticeable changes in the boiling point, and even the melting point can be affected slightly by impurities, making them unsuitable as precise standard fixed points.

(b) On the Kelvin absolute scale, one fixed point is the triple-point of water, assigned the value 273.16 K. The other fixed point is absolute zero, which corresponds to the lowest possible temperature at which the thermal motion of particles is minimum. This point is defined as 0 K on the Kelvin scale.

(c) The relation between the Celsius temperature \(t_c\) and the absolute temperature \(T\) is given by $$t_c = T - 273.15.$$ Although the triple-point of water is 273.16 K, the value 273.15 appears in this relation because the Celsius scale was historically defined using the melting point of ice as 0 °C, which corresponds to 273.15 K on the Kelvin scale. The difference of 0.01 K arises from the fact that the triple-point of water is slightly higher than the melting point of ice.

(d) On an absolute temperature scale whose unit interval is equal in size to that of the Fahrenheit scale, the zero of the scale still corresponds to absolute zero. Since one kelvin corresponds to 1.8 degrees on the Fahrenheit scale, the temperature of the triple-point of water on this absolute Fahrenheit-based scale is $$273.16 \times 1.8 \approx 491.7.$$ Thus, the triple-point of water corresponds to approximately 491.7 degrees on this absolute scale.

Q5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Solution

For an ideal gas thermometer operated at constant volume, the absolute temperature is directly proportional to the pressure of the gas. Hence, the temperature \(T\) corresponding to any pressure \(P\) is given by $$ T = T_{\text{tp}} \frac{P}{P_{\text{tp}}}, $$ where \(T_{\text{tp}} = 273.16\,\text{K}\) is the triple-point temperature of water, \(P_{\text{tp}}\) is the pressure at the triple-point, and \(P\) is the pressure at the required temperature.

For thermometer A (oxygen), the pressure at the triple-point of water is \(P_{\text{tp,A}} = 1.250 \times 10^{5}\,\text{Pa}\) and the pressure at the normal melting point of sulphur is \(P_{\text{A}} = 1.797 \times 10^{5}\,\text{Pa}\). The absolute temperature of the melting point of sulphur as read by thermometer A is therefore $$ \begin{aligned} T_{\text{A}} &= 273.16 \times \frac{1.797 \times 10^{5}}{1.250 \times 10^{5}} \\ &= 273.16 \times 1.4376 \\ &\approx 392.7\,\text{K}. \end{aligned} $$

For thermometer B (hydrogen), the pressure at the triple-point of water is \(P_{\text{tp,B}} = 0.200 \times 10^{5}\,\text{Pa}\) and the pressure at the normal melting point of sulphur is \(P_{\text{B}} = 0.287 \times 10^{5}\,\text{Pa}\). The corresponding absolute temperature is $$ \begin{aligned} T_{\text{B}} &= 273.16 \times \frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \\ &= 273.16 \times 1.435 \\ &\approx 392.1\,\text{K}. \end{aligned} $$

Thus, the absolute temperature of the normal melting point of sulphur is approximately \(392.7\,\text{K}\) according to thermometer A and \(392.1\,\text{K}\) according to thermometer B.

The slight difference in the readings arises because real gases do not behave as perfectly ideal gases at finite pressures. Oxygen and hydrogen deviate from ideal behaviour by different amounts, leading to small discrepancies in the measured temperatures even though the thermometers themselves are not faulty. To reduce this discrepancy, the experiment should be repeated at progressively lower gas pressures and the readings extrapolated to zero pressure, where all gases approach ideal behaviour and yield the same thermodynamic temperature.

Q6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = \(\mathrm{1.20 × 10^{–5}\ K^{–1}}\) .

Solution

The steel tape is correctly calibrated at \(27.0^\circ\text{C}\). On a hot day at \(45.0^\circ\text{C}\), the tape expands, so each centimetre marking on the tape becomes slightly longer. As a result, the measured length of the steel rod appears smaller than its actual length.

The length reading observed at \(45.0^\circ\text{C}\) is \(63.0\ \text{cm}\). Let the actual length of the rod on that day be \(L\). The relation between the apparent length and the true length is $$ \begin{aligned} L_{\text{apparent}} &= \frac{L}{1+\alpha (T-T_0)} . \end{aligned} $$

Substituting the given values \(\alpha = 1.20\times10^{-5}\ \text{K}^{-1}\), \(T=45^\circ\text{C}\), and \(T_0=27^\circ\text{C}\), $$ \begin{aligned} 63.0 &= \frac{L}{1+1.2\times10^{-5}(45-27)} \\ 63.0 &= \frac{L}{1+1.2\times10^{-5}\times18} \\ 63.0 &= \frac{L}{1.000216}. \end{aligned} $$

Solving for \(L\), $$ \begin{aligned} L &= 63.0 \times 1.000216 \\ &\approx 63.014\ \text{cm}. \end{aligned} $$ Thus, the actual length of the steel rod on the hot day is approximately \(63.014\ \text{cm}\).

To find the le

Q7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : \(α_{steel}= 1.20 × 10^{–5}\ K^{–1}\).

Solution

At 27 °C, the outer diameter of the steel shaft is 8.70 cm, while the diameter of the central hole of the wheel is 8.69 cm. For the wheel to just slip onto the shaft, the shaft must contract so that its diameter becomes equal to 8.69 cm.

The linear expansion (or contraction) relation for a dimension is given by

\[ L = L_0 \left( 1 + \alpha \Delta T \right) \]

Here, \(L_0 = 8.70\ \text{cm}\) is the original diameter of the shaft at 27 °C, \(L = 8.69\ \text{cm}\) is the required diameter, \(\alpha = 1.20 \times 10^{-5}\ \text{K}^{-1}\), and \(\Delta T = (t - 27)\).

\[ \begin{aligned} 8.69 &= 8.70 \left[ 1 + \alpha (t - 27) \right] \\ \frac{8.69}{8.70} &= 1 + \alpha (t - 27) \\ \alpha (t - 27) &= \frac{8.69 - 8.70}{8.70} \\ t - 27 &= \frac{-0.01}{8.70 \times 1.20 \times 10^{-5}} \\ t &= 27 - \frac{0.01}{8.70 \times 1.20 \times 10^{-5}} \\ t &\approx 27 - 95.8 \\ t &\approx -68.8^\circ\text{C} \end{aligned} \]

Therefore, the shaft must be cooled to approximately \(-69^\circ\text{C}\) for the wheel to slip onto it.

Q8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = \(\mathrm{1.70 × 10^{–5}\ K^{–1}}\).

Solution

The diameter of the circular hole at 27 °C is 4.24 cm. When the copper sheet is heated, the hole also expands in the same proportion as the material, since thermal expansion does not depend on whether the dimension corresponds to material or a cavity.

The linear expansion relation for the diameter is

\[ L = L_0 \left[ 1 + \alpha \Delta T \right] \]

Here, \(L_0 = 4.24\ \text{cm}\), \(\alpha = 1.70 \times 10^{-5}\ \text{K}^{-1}\), and \(\Delta T = 227 - 27 = 200\ \text{K}\).

\[ \begin{aligned} L &= 4.24 \left[ 1 + 1.70 \times 10^{-5} (227 - 27) \right] \\ &= 4.24 \left[ 1 + 1.70 \times 10^{-5} \times 200 \right] \\ &= 4.24 \left( 1 + 0.0034 \right) \\ &= 4.254\ \text{cm} \end{aligned} \]

The change in the diameter of the hole is therefore

\[ \begin{aligned} \Delta L &= 4.254 - 4.24 \\ &= 1.44 \times 10^{-2}\ \text{cm} \end{aligned} \]

Hence, on heating the copper sheet to 227 °C, the diameter of the hole increases by \(1.44 \times 10^{-2}\ \text{cm}\).

Q9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = \(\mathrm{2.0 × 10^{–5}\ K^{–1}}\); Young’s modulus of brass = \(\mathrm{0.91 × 10^{11}}\) Pa.

Solution

The length of the brass wire at 27 °C is 1.8 m. The wire is held between rigid supports, so on cooling it cannot contract freely. As a result, thermal contraction produces tensile stress in the wire. The temperature is reduced from 27 °C to –39 °C, giving a temperature change of 66 K.

The fractional change in length due to thermal contraction is given by

\[ \frac{\Delta L}{L} = \alpha \Delta T \]

\[ \begin{aligned} \frac{\Delta L}{1.8} &= 2.0 \times 10^{-5} \times \left[ 27 - (-39) \right] \\ &= 2.0 \times 10^{-5} \times 66 \\ \Delta L &= 1.8 \times 2.0 \times 10^{-5} \times 66 \\ &= 2.376 \times 10^{-3}\ \text{m} \end{aligned} \]

Since the wire is constrained, this contraction produces tensile strain. Using Young’s modulus,

\[ Y = \frac{\sigma}{\varepsilon} \]

\[ \begin{aligned} \sigma &= Y \varepsilon \\ &= 0.91 \times 10^{11} \times \frac{\Delta L}{L} \\ &= 0.91 \times 10^{11} \times \frac{2.376 \times 10^{-3}}{1.8} \\ &\approx 1.20 \times 10^{8}\ \text{Pa} \end{aligned} \]

The radius of the wire is \(1.0 \times 10^{-3}\ \text{m}\). Hence, the cross-sectional area is

\[ A = \pi r^2 = \pi \left( 1.0 \times 10^{-3} \right)^2 \]

\[ \begin{aligned} F &= \sigma A \\ &= 1.20 \times 10^{8} \times \pi \left( 1.0 \times 10^{-3} \right)^2 \\ &\approx 3.8 \times 10^{2}\ \text{N} \end{aligned} \]

Therefore, the tension developed in the brass wire on cooling is approximately \(3.8 \times 10^{2}\ \text{N}\).

Q10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = \(\mathrm{2.0 × 10^{–5}\ K^{–1}}\), steel = \(\mathrm{1.2 × 10^{–5}\ K^{–1}}\) ).

Solution

The length of each rod at the initial temperature of 40 °C is 50 cm. The brass rod and the steel rod are rigidly joined end to end, but the combined rod has free ends. When the temperature is raised to 250 °C, each rod is free to expand according to its own coefficient of linear expansion. The temperature rise is therefore \(\Delta T = 250 - 40 = 210\ \text{K}\).

The length of the brass rod after heating is obtained from the linear expansion relation

\[ L_b = L_{b0}\left(1 + \alpha_b \Delta T\right) \]

\[ \begin{aligned} L_b &= 50 \left[ 1 + 2.0 \times 10^{-5} \times 210 \right] \\ &= 50 \left( 1 + 4.2 \times 10^{-3} \right) \\ &= 50.21\ \text{cm} \end{aligned} \]

The length of the steel rod after heating is

\[ L_s = L_{s0}\left(1 + \alpha_s \Delta T\right) \]

\[ \begin{aligned} L_s &= 50 \left[ 1 + 1.2 \times 10^{-5} \times 210 \right] \\ &= 50 \left( 1 + 2.52 \times 10^{-3} \right) \\ &= 50.126\ \text{cm} \end{aligned} \]

The combined length of the brass–steel rod at 250 °C is therefore

\[ \begin{aligned} L_c &= L_b + L_s \\ &= 50.21 + 50.126 \\ &= 100.336\ \text{cm} \end{aligned} \]

Since the original combined length at 40 °C was 100 cm, the total increase in length is \(0.336\ \text{cm}\). As the ends of the rod are free to expand, no thermal stress is developed at the junction between the brass and steel rods.

Q11. The coefficient of volume expansion of glycerine is \(\mathrm{49 × 10^{–5}\ K^{–1}}\). What is the fractional change in its density for a 30 °C rise in temperature ?

Solution

The coefficient of volume expansion of glycerine is given as \(\alpha_v = 49 \times 10^{-5}\ \text{K}^{-1}\). The temperature rise is 30 °C. The fractional change in volume due to heating is obtained from the relation

\[ \frac{\Delta V}{V} = \alpha_v \Delta T \]

\[ \begin{aligned} \frac{\Delta V}{V} &= 49 \times 10^{-5} \times 30 \\ &= 1470 \times 10^{-5} \\ &= 0.0147 \end{aligned} \]

Since the mass of glycerine remains constant, density varies inversely with volume. Therefore, the fractional change in density is equal in magnitude but opposite in sign to the fractional change in volume.

\[ \begin{aligned} \frac{\Delta \rho}{\rho} &= -\frac{\Delta V}{V} \\ &= -0.0147 \end{aligned} \]

Hence, for a temperature rise of 30 °C, the density of glycerine decreases by 0.0147, which corresponds to a decrease of approximately 1.47 %.

Q12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = \(\mathrm{0.91 J\ g^{–1}\ K^{–1}}\).

Solution

The mass of the aluminium block is 8.0 kg and the drilling time is 2.5 minutes. The power of the drilling machine is 10 kW, which is equivalent to \(10^{4}\ \text{J s}^{-1}\). The total energy supplied by the machine during this time is first calculated.

\[ \begin{aligned} Q &= 10^{4} \times 2.5 \times 60 \\ &= 1.5 \times 10^{6}\ \text{J} \end{aligned} \]

Since only 50 % of this energy is actually used to heat the aluminium block, the heat gained by the block is

\[ \begin{aligned} Q_{\text{useful}} &= 1.5 \times 10^{6} \times \frac{50}{100} \\ &= 7.5 \times 10^{5}\ \text{J} \end{aligned} \]

The specific heat capacity of aluminium is \(0.91\ \text{J g}^{-1}\ \text{K}^{-1}\). Converting the mass into grams, \(m = 8.0 \times 10^{3}\ \text{g}\). Using the heat equation,

\[ Q = mc\Delta T \]

\[ \begin{aligned} 7.5 \times 10^{5} &= 8.0 \times 10^{3} \times 0.91 \times \Delta T \\ \Delta T &= \frac{7.5 \times 10^{5}}{8.0 \times 10^{3} \times 0.91} \\ &\approx 1.03 \times 10^{2}\ \text{K} \end{aligned} \]

Hence, the rise in temperature of the aluminium block in 2.5 minutes is approximately \(103\ \text{K}\) (or 103 °C).

Q13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = \(\mathrm{0.39\ J\ g^{–1}\ K^{–1}}\); heat of fusion of water = \(\mathrm{335\ J\ g^{–1}}\) ).

Solution

The mass of the copper block is \(2.5\,\text{kg}\), which is equal to \(2.5 \times 10^{3}\,\text{g}\). The copper block is heated to \(500^\circ\text{C}\) and then placed on a large ice block. Since the ice is initially at \(0^\circ\text{C}\), the copper cools down from \(500^\circ\text{C}\) to \(0^\circ\text{C}\). The specific heat capacity of copper is \(0.39\,\text{J g}^{-1}\text{K}^{-1}\) and the heat of fusion of ice is \(335\,\text{J g}^{-1}\).

The heat lost by the copper block is given by

\[ \begin{aligned} Q_{\text{Cu}} &= mc\Delta T \\ &= (2.5 \times 10^{3}) \times 0.39 \times 500 \\ &= 4.875 \times 10^{5}\,\text{J}. \end{aligned} \]

This heat is used entirely to melt ice. If \(m\) is the mass of ice melted, then the heat gained by the ice is

\[ \begin{aligned} Q_{\text{ice}} &= mL \\ &= m \times 335. \end{aligned} \]

Equating the heat lost by copper to the heat gained by ice,

\[ \begin{aligned} 4.875 \times 10^{5} &= m \times 335 \\ m &= \frac{4.875 \times 10^{5}}{335} \\ &= 1.45 \times 10^{3}\,\text{g} \\ &= 1.45\,\text{kg}. \end{aligned} \]

Therefore, the maximum amount of ice that can melt is approximately \(1.45\,\text{kg}\).

Q14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?

Solution

The mass of the metal block is \(0.20\,\text{kg}\) and its initial temperature is \(150^\circ\text{C}\). It is dropped into a copper calorimeter having a water equivalent of \(0.025\,\text{kg}\), which contains \(150\,\text{cm}^3\) of water at \(27^\circ\text{C}\). The final equilibrium temperature of the system is \(40^\circ\text{C}\). Let the specific heat of the metal be \(c\).

The heat lost by the hot metal is equal to the heat gained by the water and the calorimeter.

The heat lost by the metal is

\[ \begin{aligned} Q_{\text{metal}} &= mc\Delta T \\ &= 0.2 \times 10^{3} \times c \times (150 - 40) \\ &= 0.2 \times 10^{3} \times 110 \times c \\ &= 2.2 \times 10^{4} c . \end{aligned} \]

The calorimeter has a water equivalent of \(0.025\,\text{kg}\). The heat gained by the calorimeter is

\[ \begin{aligned} Q_{\text{cal}} &= mc\Delta T \\ &= 0.025 \times 4186 \times (40 - 27) \\ &\approx 1.36 \times 10^{3}\,\text{J}. \end{aligned} \]

The volume of water is \(150\,\text{cm}^3\). Taking the density of water as \(1\,\text{g cm}^{-3}\), the mass of water is

\[ \begin{aligned} m &= \rho V \\ &= 1 \times 150 \\ &= 150\,\text{g} = 0.150\,\text{kg}. \end{aligned} \]

The heat gained by the water is

\[ \begin{aligned} Q_{\text{w}} &= mc\Delta T \\ &= 0.150 \times 4186 \times (40 - 27) \\ &\approx 8.16 \times 10^{3}\,\text{J}. \end{aligned} \]

Equating heat lost by the metal to the total heat gained by the water and the calorimeter,

\[ \begin{aligned} 2.2 \times 10^{4} c &= 1.36 \times 10^{3} + 8.16 \times 10^{3} \\ &= 9.52 \times 10^{3} \\ c &= \frac{9.52 \times 10^{3}}{2.2 \times 10^{4}} \\ &\approx 0.433\,\text{J g}^{-1}\text{K}^{-1}. \end{aligned} \]

Thus, the specific heat of the metal is approximately \(0.43\,\text{J g}^{-1}\text{K}^{-1}\) (or \(430\,\text{J kg}^{-1}\text{K}^{-1}\)). If heat losses to the surroundings are not negligible, some heat lost by the metal is not accounted for in heating the water and calorimeter. As a result, the calculated value of specific heat will be smaller than the actual value.

Q15. Given below are observations on molar specific heats at room temperature of some common gases.

Solution

For an ideal monatomic gas, the atoms can store energy only in translational motion along three mutually perpendicular directions. According to the principle of equipartition of energy, each translational degree of freedom contributes \(\tfrac{1}{2}R\) per mole to the molar specific heat at constant volume. Hence, for a monatomic gas,

\[ \begin{aligned} C_v &= \frac{3}{2}R \\ &= \frac{3}{2} \times 1.987 \\ &\approx 2.98 \,\text{cal mol}^{-1}\text{K}^{-1}, \end{aligned} \]

which is close to the experimentally observed value of about \(2.92\,\text{cal mol}^{-1}\text{K}^{-1}\).

The gases listed in the table are diatomic molecules. In addition to translational motion, a diatomic molecule can also rotate about two axes perpendicular to the line joining the atoms. At ordinary temperatures, vibrational modes are usually not excited, but the two rotational degrees of freedom are active. Therefore, a diatomic gas has a total of five active degrees of freedom.

Using the equipartition theorem, the molar specific heat at constant volume for a diatomic gas is

\[ \begin{aligned} C_v &= \frac{5}{2}R \\ &= \frac{5}{2} \times 1.987 \\ &\approx 4.97 \,\text{cal mol}^{-1}\text{K}^{-1}. \end{aligned} \]

This theoretical value agrees well with the observed molar specific heats of hydrogen, nitrogen, oxygen, nitric oxide, and carbon monoxide, all of which lie close to \(5\,\text{cal mol}^{-1}\text{K}^{-1}\). The marked difference from the monatomic value arises due to the additional rotational degrees of freedom present in diatomic molecules.

Chlorine, however, has a noticeably larger molar specific heat. This indicates that, at room temperature, some vibrational modes of the chlorine molecule are also partially excited. The contribution of vibrational energy increases the internal energy change per unit temperature rise, leading to a higher value of \(C_v\). Thus, the comparatively large molar specific heat of chlorine suggests the onset of vibrational degrees of freedom in addition to translational and rotational motions.

Q16. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1

Solution

The initial temperature of the child is \(101^\circ\text{F}\) and the final temperature is \(98^\circ\text{F}\). Thus, the fall in temperature is \(3^\circ\text{F}\). The mass of the child is \(30\,\text{kg}\). The specific heat of the human body is taken to be the same as that of water, i.e. \(1\,\text{cal g}^{-1}\text{K}^{-1}\). The latent heat of evaporation of water is \(580\,\text{cal g}^{-1}\), and the time taken is \(20\,\text{minutes}\).

First, the fall in temperature is converted from Fahrenheit to Celsius. Since a change of \(180^\circ\text{F}\) corresponds to a change of \(100^\circ\text{C}\),

\[ \begin{aligned} 3^\circ\text{F} &= \frac{100}{180} \times 3 \\ &= 1.67^\circ\text{C}. \end{aligned} \]

The heat lost by the body due to this fall in temperature is

\[ \begin{aligned} Q &= mc\Delta T \\ &= 30 \times 10^{3} \times 1 \times 1.67 \\ &= 5.01 \times 10^{4}\,\text{cal}. \end{aligned} \]

This heat is assumed to be lost entirely due to evaporation of sweat. If \(m\) is the mass of water evaporated, then

\[ \begin{aligned} Q &= mL \\ 5.01 \times 10^{4} &= m \times 580 \\ m &= \frac{5.01 \times 10^{4}}{580} \\ &\approx 86.38\,\text{g}. \end{aligned} \]

The average rate of extra evaporation caused by the drug is therefore

\[ \begin{aligned} \text{Rate of evaporation} &= \frac{m}{t} \\ &= \frac{86.38}{20} \\ &\approx 4.32\,\text{g min}^{-1}. \end{aligned} \]

Hence, the average rate of extra evaporation caused by the antipyrin is approximately \(4.3\,\text{g min}^{-1}\).

Q17. A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is \(\mathrm{0.01\ J\ s^{–1}\ m^{–1}\ K^{–1}}\). [Heat of fusion of water = \(\mathrm{335 × 103\ J\ kg^{–1}}\)

Solution

The icebox is cubical in shape with an inner side length of \(30\,\text{cm}\). The thickness of the thermacole walls is \(5.0\,\text{cm}\). Hence, the outer side length of the cube is

\[ \begin{aligned} L_{\text{outer}} &= 30 + 2 \times 5 \\ &= 40\,\text{cm} = 0.40\,\text{m}. \end{aligned} \]

The inner side length is \(0.30\,\text{m}\). The area through which heat flows is the total outer surface area of the cube,

\[ \begin{aligned} A &= 6L_{\text{outer}}^2 \\ &= 6 \times (0.40)^2 \\ &= 0.96\,\text{m}^2. \end{aligned} \]

The thickness of the thermacole wall is

\[ \begin{aligned} x &= 5.0\,\text{cm} = 0.05\,\text{m}. \end{aligned} \]

The temperature inside the icebox is approximately \(0^\circ\text{C}\), since ice is present, while the outside temperature is \(45^\circ\text{C}\). Thus, the temperature difference across the wall is

\[ \begin{aligned} \Delta T &= 45 - 0 = 45\,\text{K}. \end{aligned} \]

The rate of heat flow into the icebox by conduction is given by

\[ \begin{aligned} \frac{Q}{t} &= \frac{kA\Delta T}{x} \\ &= \frac{0.01 \times 0.96 \times 45}{0.05} \\ &= 8.64\,\text{J s}^{-1}. \end{aligned} \]

The total time for which heat flows into the box is \(6\,\text{h} = 6 \times 3600 = 21600\,\text{s}\). Hence, the total heat entering the icebox is

\[ \begin{aligned} Q &= 8.64 \times 21600 \\ &= 1.87 \times 10^{5}\,\text{J}. \end{aligned} \]

This heat is used to melt the ice. If \(m\) is the mass of ice melted and the heat of fusion of ice is \(335 \times 10^{3}\,\text{J kg}^{-1}\), then

\[ \begin{aligned} Q &= mL \\ 1.87 \times 10^{5} &= m \times 335 \times 10^{3} \\ m &= 0.56\,\text{kg}. \end{aligned} \]

Initially, the mass of ice in the box is \(4.0\,\text{kg}\). Therefore, the mass of ice remaining after \(6\,\text{h}\) is

\[ \begin{aligned} m_{\text{remaining}} &= 4.0 - 0.56 \\ &= 3.44\,\text{kg}. \end{aligned} \]

Hence, approximately \(3.4\,\text{kg}\) of ice remains in the thermacole icebox after six hours.

Q18. A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = \(\mathrm{109\ J\ s^{–1}\ m{–1}\ K^{–1}}\) ; Heat of vaporisation of water = \(\mathrm{2256 × 10^3\ J\ kg^{–1}}\).

Solution

The base area of the brass boiler is \(A = 0.15\,\text{m}^2\) and
its thickness is \(x = 1.0\,\text{cm} = 0.01\,\text{m}\).
Water boils at a rate of \(6.0\,\text{kg min}^{-1}\), which is equal to \(0.10\,\text{kg s}^{-1}\).
The thermal conductivity of brass is \(k = 109\,\text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}\), and
the heat of vaporisation of water is \(L = 2.256 \times 10^{6}\,\text{J kg}^{-1}\).

The heat supplied per second to convert water into steam is

\[ \begin{aligned} \frac{Q}{t} &= mL \\ &= 0.10 \times 2.256 \times 10^{6} \\ &= 2.256 \times 10^{5}\,\text{J s}^{-1}. \end{aligned} \]

This heat is conducted through the brass base of the boiler. If \(T_f\) is the temperature of the flame in contact with the boiler and the inner surface of the boiler is at \(100^\circ\text{C}\), then the rate of heat conduction is

\[ \begin{aligned} \frac{Q}{t} &= \frac{kA(T_f - 100)}{x}. \end{aligned} \]

Substituting the given values,

\[ \begin{aligned} 2.256 \times 10^{5} &= \frac{109 \times 0.15 \times (T_f - 100)}{0.01}. \end{aligned} \]

Solving for \(T_f\),

\[ \begin{aligned} T_f - 100 &= \frac{2.256 \times 10^{5} \times 0.01}{109 \times 0.15} \\ &\approx 138 \\ T_f &\approx 238^\circ\text{C}. \end{aligned} \]

Hence, the temperature of the part of the flame in contact with the boiler is approximately \(2.4 \times 10^{2}\,^\circ\text{C}\).

Q19. Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Solution

A body with large reflectivity reflects most of the radiation incident on it and absorbs very little. Since, at a given temperature, good absorbers are also good emitters as stated by Kirchhoff’s law of thermal radiation, a body that absorbs little energy must also emit little energy. Hence, a body with large reflectivity behaves as a poor emitter of thermal radiation.

On a chilly day, a brass tumbler feels much colder than a wooden tray even though both are at the same temperature as the surroundings. Brass is a good conductor of heat, so it draws heat rapidly from the hand, lowering the temperature of the skin quickly and creating a strong sensation of cold. Wood, being a poor conductor, removes heat from the hand much more slowly, so it does not feel as cold.

An optical pyrometer measures temperature by comparing the intensity of radiation emitted by a hot body with that of an ideal black body at the same wavelength. A red hot iron piece in the open is not a perfect black body and has emissivity less than unity, so it emits less radiation than an ideal black body at the same temperature. As a result, the pyrometer interprets this lower intensity as a lower temperature. When the same iron piece is placed in a furnace, it effectively behaves like a black body due to multiple reflections and absorption of radiation within the enclosure, and hence the pyrometer gives the correct temperature.

If the earth had no atmosphere, it would lose a large amount of heat directly into space by radiation, especially during the night. The atmosphere absorbs and re-radiates part of the outgoing infrared radiation back to the earth’s surface, thereby reducing heat loss and maintaining a moderate temperature. In the absence of this greenhouse effect, the earth would become extremely cold and inhospitable to life.

Q20. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Solution

According to Newton’s law of cooling, the rate of fall of temperature of a body is proportional to the excess of its temperature over that of the surroundings. This implies that the temperature difference with the surroundings decreases exponentially with time.

Let the surrounding temperature be \(T_s = 20^\circ\text{C}\). If \(T\) is the temperature of the body at time \(t\), then

\[ \begin{aligned} T - T_s &= (T_0 - T_s)e^{-kt}, \end{aligned} \]

where \(T_0\) is the initial temperature and \(k\) is a constant for the body and the surrounding conditions.

The body cools from \(80^\circ\text{C}\) to \(50^\circ\text{C}\) in \(5\) minutes. Substituting these values,

\[ \begin{aligned} 50 - 20 &= (80 - 20)e^{-5k} \\ 30 &= 60e^{-5k} \\ e^{-5k} &= \frac{1}{2}. \end{aligned} \]

Taking natural logarithms,

\[ \begin{aligned} -5k &= \ln\!\left(\frac{1}{2}\right) \\ k &= \frac{\ln 2}{5}. \end{aligned} \]

Now, let \(t\) be the time taken for the body to cool from \(60^\circ\text{C}\) to \(30^\circ\text{C}\). Using the same relation,

\[ \begin{aligned} 30 - 20 &= (60 - 20)e^{-kt} \\ 10 &= 40e^{-kt} \\ e^{-kt} &= \frac{1}{4}. \end{aligned} \]

Taking natural logarithms again,

\[ \begin{aligned} -kt &= \ln\!\left(\frac{1}{4}\right) = -\ln 4 \\ t &= \frac{\ln 4}{k}. \end{aligned} \]

Substituting \(k = \frac{\ln 2}{5}\),

\[ \begin{aligned} t &= \frac{\ln 4}{\ln 2/5} \\ &= \frac{2\ln 2}{\ln 2/5} \\ &= 10\,\text{minutes}. \end{aligned} \]

Therefore, the time taken by the body to cool from \(60^\circ\text{C}\) to \(30^\circ\text{C}\) is \(10\) minutes.