Question

The temperature of a 0.700 kg cube of ice is decreased

to -150 C. Then energy is gradually transferred to the cube as

heat while it is otherwise thermally

isolated from its environment. The

total transfer is 0.6993 MJ. Assume

the value of c for ice given

is valid for temperatures from

-150 C to 0 C. What is the final

temperature of the water?

Answer 1

Let $m = 0.7kg$ be the mass of ice, $c_i = 2093J/kg/\degree C$ specific heat of ice, $\lambda = 333\times 10^3 J/kg$ heat of fusion of ice, $T_i = -150\degree C$ the initial temperature of ice, $Q = 0.6993\times 10^{6}J$ is the supplied heat.

In order to raise the temperature of ice from $T_i$ to 0, the following amount of heat is required:

$$Q_{heat} = c_im(0\degree C-T_i)= -c_imT_i$$

In order to melt the ice, the following amount of heat is required:

$$Q_{melt} = \lambda m$$

The remaining heat:

$$Q - Q_{heat}-Q_{melt}$$

will heat up the water from $0\degree C$ to the temperature $T_f$. Thus, obtain:

$$Q - Q_{heat}-Q_{melt} = c_wmT_f$$

where $c_w = 4200J/kg/\degree C$ is the specific heat of water.

Expressing $T_f$, obtain:

$$T_f = \dfrac{Q - Q_{heat}-Q_{melt}}{c_w m} = \dfrac{Q +c_imT_i- \lambda m}{c_wm}\\ T_f = \dfrac{0.6993\times 10^{6} - 2093\cdot 0.7\cdot 150 - 333\times 10^3\cdot 0.7}{4200\cdot 0.7} \approx 83.8\degree C$$

Answer. $83.8\degree C$.