The temperature of a$\mathbf{0}\mathbf{.700}\mathbf{}\mathbf{\text{kg}}$cube of ice is decreased to$\mathbf{-}\mathbf{150}\mathbf{°}\mathbf{\text{C}}$. Then energy is gradually transferred to the cube as heat while it is otherwise thermally isolated from its environment. The total transfer is$\mathbf{0}\mathbf{.6993}\mathbf{}\mathbf{\text{\hspace{0.17em}MJ}}$. Assume the value of${\mathit{c}}_{\mathbf{i}\mathbf{c}\mathbf{e}}$given in Table 18-3 is valid for temperatures from$\mathbf{-}\mathbf{150}\mathbf{}\mathbf{°}{\mathbf{\text{C\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}0}}}^{\mathbf{\text{0}}}\mathbf{\text{C}}$. What is the final temperature of the water?

1. Mass of the ice cube is$m=0.700\text{kg}$
2. Initial temperature is$T_i=-{150}^0\text{C}$
3. Total heat transfer is .$Q=0.6993\text{MJ}$

Conduction heat transfer is the transfer of heat through matter without the bulk motion of the matter. This process involves the transfer of the high-energy particle to the end of a low energetic particle, when two bodies with different temperatures are brought in contact inside a system. First, the heat is used to raise the temperature and the heat is used to change the phase from ice to liquid, and then, the remaining heat is used to raise the temperature.

Formulae:

The heat transferred by the body due to thermal radiation,$Q=mc\Delta T$…(i)

where m is the mass of the substance,is the specific heat of the substance, and$\Delta T$ is the temperature difference.

The heat released due to the thermal radiation by the body $Q=mL$,…(ii)

where $L_f$is the latent heat of fusion and is the mass of the substance.

The heat given is used to raise the temperature from $-{150}^{0}to{0}^0$, and the heat is used to change the phase from ice to liquid, and then the remaining heat is used to raise temperature from$0^{0}to{T}_f$.

Using equations (i) and (ii), the total heat transferred by the ice and the liquid and heat released by the fusion process is

$Q=m{c}_{ice}\Delta T+m{L}_f+m{c}_w\Delta T$ …(iii)

where$c_{ice}$ is the specific heat of ice and is specific heat of water.

Substituting the given values in equation (iii), we get the final temperature as

$\begin{array}{c}0.6993\times {10}^6\text{J}=\left(0.700\text{kg}\right)\left(2220{\text{J/kg.}}^0\text{C}\right)(0^0\text{C}-(-{150}^0\text{C}))+\left(0.700\text{kg}\right))(333\times {10}^3\text{J/kg})\\ +\left(0.700\text{kg}\right)\left(4187{\text{J/kg.}}^0\text{C}\right)(T_f-0^0\text{C})\\ 0.6993\times {10}^6\text{J}=466200\text{J}+2930.9(T_f-0^0\text{C})\\ 233100\text{J}=\left({\text{2930.9J/}}^0\text{C}\right)T_f\\ T_f={79.5}^0\text{C}\end{array}$

Hence, the final temperature is${79.5}^0\text{C}$