Question

A mass of 25.5 g of H2O(g) at 373 K is mixed with 325 g of H2O(l) at 285 K and 1 atm. Calculate the final temperature of the system once equilibrium has been reached. Assume that ????P,m for H2O is constant at its values for 298 K throughout the temperature range of interest.

Answer 1

Answer:

331.28 K

Explanation:

To solve this problem, you need to know that the heat that the water at 373 K is equal to the heat that the water at 285 K gains.

First, we will asume that at the end of this process there won't be any water left in gaseous state.

The heat that the steam (H20(g)) loses is equal to the heat lost because the change of phase plus the heat lost because of the decrease in temperature:

$Q_g = c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg})$

The specific Heat c of water at 298K is 4.18 kJ/K*kg.

The latent heat cl of water is equal to  2257 kJ/kg.

The heat that the cold water gains is equal to heat necessary to increase its temperature to its final value:

$Q_l = m_{wl}*c*(T_{fl}-T_{ol})$

Remember that in equilibrium, the final temperature of both bodies of water will be equal.

Then:

$Q_g = Q_l\\c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg}) = m_{wl}*c*(T_{fl}-T_{ol})\\2257 kJ/kg*0.0255 kg + 0.0255 kg*4.18 kJ/kg*K*(373K - T_f) = 0.325 kg*4.18kJ/kg*K*(T_f-285K)\\57.5535 kJ + 39.75807kJ - 0.10659T_f = 1.3585 T_f - 387.1725 kJ\\484.48407 kJ = 1.46244 T_f\\T_f = 484.48407 kJ /1.46244 = 331.28 K$