Question 6.2: Water is cooled with ice cubes (Fig. 6.10). The water and th...

Water is cooled with ice cubes (Fig. 6.10). The water and the ice cubes are considered as an isolated system. Initially, the ice cubes are at melting temperature T_0 and the water at temperature T_i. The total initial mass of ice is \acute{M} and the initial mass of water is M. The latent heat of melting of ice per unit mass is \ell^*_{s\ell} and the specific heat per unit mass of water is c^∗_ V.

a) Determine the final temperature T_m of the water.
b) Determine the final temperature T_m of the water if melted ice (i.e. water) had been added at melting temperature T_0 instead of ice.

Numerical Application:

M = 0.45 kg, \acute{M} = 0.05 kg, T_i = 20°C, T_0 = 0°C, \ell^*_{s\ell} = 3.33 · 10^6 J kg^{−1} and c^∗_V = 4.19 · 10^3 J kg^{−1} K^{−1}.

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Heat is transferred from the water to melt the ice and increases its temperature while the temperature of the water decreases.

a) Since the system is isolated the total variation of internal energy vanishes,

ΔU_{if} = \acute{M} \ell^*_{s f} + \acute{M}c^*_V (T_f – T_0) +M c^*_V (T_f -T_i ) =0 .

Thus, the final temperature T_f is,

T_f = \frac{M T_i + \acute{M} T_0 }{M + \acute{M} } – \frac{\acute{M} \ell^* _{sf} }{(M+\acute{M} )c^*_V} = 283 K = 10^\circ C .

b) If the melted ice had been added at melting temperature T_0 instead of ice, there would be no more melting. Thus, the final temperature T_f would be,

T_f = \frac{M T_i + \acute{M} T_0 }{M + \acute{M} } = 291 K = 18° C .

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