Question

Equal masses of two different liquids have the same temperature of

Answer 1

The full condition of the problem is:

Equal masses of two different liquids have the same temperature of 25.5°C. Liquid A has a freezing point of -66.3°C and a specific heat capacity of 1770 J/(kg C°). Liquid B has a freezing point of -89.3°C and a specific heat capacity of 2630 J/(kg C°). The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference Lf,A - Lf,B between the latent heats of fusion for these liquids.

Solution

Heat to remove from the first and second liquid, respectively:

$$Q=c_Am\Delta t_A+mL_{fA},\\ Q=c_Bm\Delta t_B+mL_{fB}.\\\space\\ c_Am\Delta t_A+mL_{fA}=c_Bm\Delta t_B+mL_{fB},\\\space\\ L_{fA}-L_{fB}=c_B\Delta t_B-c_A\Delta t_A,\\ L_{fA}-L_{fB}=\\=2630(-89.3-25.5)-1770(-66.3-25.5)=\\ =-139438\text{ J/kg}.$$