A 0.6-kg block of ice at -10 °C is placed into a closed empty 1 m^3 container, also at a temperature of -10 °C. The temperature of the container is then increased to 100 °C. How much greater is the heat required than that necessary to raise the empty container alone to that temperature?

Question

A [latex] 0.6-kg [/latex] block of ice at [latex] -10 °C [/latex] is placed into a closed empty [latex] 1 m^3[/latex] container, also at a temperature of [latex] -10 °C.[/latex] The temperature of the container is then increased to [latex] 100 °C.[/latex] How much greater is the heat required than that necessary to raise the empty container alone to that temperature?

Accepted Answer

Consider first the state of the ice and the pressure in the container at [latex]100°C.[/latex] Since the density of saturated water vapour is [latex] 0.5977(≈ 0.6) kg \ m^{-3} [/latex] and its pressure is [latex] 1 [/latex] atm, the [latex] 0.6 kg [/latex] of ice is completely transformed into vapour at [latex] 100°C,[/latex] indeed into saturated vapour at a pressure of [latex] 1 [/latex] atm.

How does ice, at a temperature of [latex] -10 °C, [/latex] turn into saturated water vapour at [latex] 100 °C[/latex] ? If the temperature is increased very slowly, the system passes through a number of equilibrium states. First, the ice sublimes and the ice phase is in equilibrium with the vapour phase. This lasts until the temperature and pressure of the triple point [latex] (0.01 °C, 610 Pa)[/latex] is reached. At the triple point, a liquid state appears alongside the ice and water vapour. Further heating makes the solid phase disappear, and only water and saturated water vapour remain in the container. It is interesting to note that subsequently the water boils steadily at [latex] 100 °C, [/latex] until all the water has been transformed into vapour.
From the point of view of heat absorption, only the initial and the final states are important. The heat [latex] Q [/latex] absorbed by the system, as it passes with increasing internal energy through the 'ice-water-vapour' states, is independent of the intermediate states. For calculational purposes, the heating of [latex] 0.6 kg [/latex] of ice should be divided into four stages (warming the ice, melting the ice, warming the water and boiling the water) to give:

[latex] Q=c_imΔT_1+L_fm+c_wmΔT_2 + L_V m, [/latex]

where [latex] c_i = 2.1 kJ \ kg^{-1} °C^{-1} [/latex] is the specific heat of ice, [latex] ΔT_1 = 10°C, L_f =334 kJ \ kg^{-1} [/latex] is the heat of fusion Of ice, [latex] c_w = 4.2 kJ \ kg^{-1} °C^{-1} [/latex] is the specific heat of water, [latex] ΔT_2 = 100 °C [/latex] and [latex] L_V [/latex] is the specific heat of vaporisation of water.

The specific heats of ice and water are slightly dependent on temperature and pressure, and the heat of fusion of the ice also depends a little on pressure. But these variations are small and can be neglected. However, the situation is different for the heat of vaporisation of water. It is true that the heat of vaporisation only varies a little with temperature, but its dependence on pressure is significant.

The value usually given in tables, [latex] L_V = 2256 kJ \ kg^{-1},[/latex] covers, not only the higher internal energy of the vapour, but also the work done against atmospheric pressure. In the present problem, this work was done when the container was evacuated, i.e. the heat to be transferred to the system is smaller by this amount, which is [latex]- pΔ V = -101.3 kJ.[/latex] As this figure is that for [latex] 0.6 kg [/latex] of water,

[latex] L_V=\left(2256-\frac{101.3}{0.6} \right) =2087\approx 2090 kJ \ kg^{-1} [/latex]

is the more accurate value to be used for the present calculation. After substitution of the data, the heat transfer is found to be [latex] Q = 1720 kJ.[/latex]

Neglecting the work done against atmospheric pressure would have introduced an error of nearly [latex] 6 [/latex] per cent, while ignoring the slight dependence of the other coefficients on temperature and pressure causes an inaccuracy of approximately [latex] 1 [/latex] per cent. The main reason for this is that the normal change in the volume in the course of a water-vapour transition is very significant (a factor of about [latex] 1600[/latex] ). The volumes of the water and ice are negligible compared with that of the vapour, and they are tacitly neglected in the solution.

Question P.140: A 0.6-kg block of ice at -10 °C is placed into a closed empt...

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