Question

Suppose an ice cube weighing 36.0 g at a temperature of $-10°C$is placed in 360 g water at a temperature of 20${}^{\circ }C$. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings. The enthalpy of fusion of ice is $\Delta H_{fus}=6.007kJmo{l}^{-1}$, and the molar heat capacities c_pof ice and water are 38 and $75J{K}^{-1}mo{l}^{-1}$, respectively.

Short answer

After reaching equilibrium, the temperature of the system will be 10.44${}^{\circ }C$.

Step-by-step solution

The given Information.

Weight of ice taken, W_ice is 36.0 g.

The temperature of ice is -10${}^{\circ }C$.

Theheat capacity of ice,$C_p=\frac{38J}{Kmol}.$.

The weight of water taken is $w_{water}=360g$.

The temperature of water, $T_{water}=20°C$.

The heat capacity of water,$C_{pwater}=\frac{75J}{Kmol}$ .

Enthalpy of fusion.

The energy that must be provided to a solid substance in order to cause a variation in the physical state and get transformed into a liquid.

This is termed the enthalpy of fusion.

The moles of ice and water.

The weight of the ice taken; W_iceis 36.0 g.

The temperature of ice is -${10}^{\circ }C$.

The molar mass of ice is $\frac{18g}{mol}$.

Hence, the moles of ice are:

$$\begin{gathered} m_{ice}=\frac{w_{ice}}{M_{ice}} \\ =\frac{36g}{18gmo{l}^{-1}} \\ =2mol \end{gathered}$$

The latent heat of fusion of ice is:

$\Delta H_{fus}=\frac{6.007kJ}{mol}or\frac{6007J}{mol}$

As the molar mass of water is $\frac{18g}{mol}$.

The number of moles of water is:

$$\begin{gathered} m_{water}=\frac{w_{water}}{M_{water}} \\ =\frac{360g}{18gmo{l}^{-1}} \\ =20mol \end{gathered}$$

Since, there is no heat loss to the surroundings, heat absorbed by ice is equal to the heat lost by water.

The temperature after thermal equilibrium is reached.

Assuming that the ice first reaches $0^{\circ }C$, then melts and gets converted into water at $0^{\circ }C$, and rises in temperature until an equilibrium temperature is reached.

As;

$\Delta H_{ice}=\Delta H_{water}$

Therefore;

$m_{ice}{C}_{pice}(0°C-T_{ice})+\left(m_{ice}\Delta H_{fus}\right)+m_{icewater}{C}_{pwater}(T_{eq}-0°C)=m_{water}{C}_{pwater}(T_{water}-T_{eq})$

Here,

T_eqis the equilibrium temperature.

$m_{ice}{C}_{pice}(0°C-T_{ice})$is the heat absorbed by the ice to reach 0C.

$m_{ice}\Delta H_{fus}$is the heat absorbed by the ice to melt into water.

$m_{icewater}{C}_{pwater}(T_{eq}-0°C)$is the heat absorbed by the melted ice (water) to reach equilibrium temperature.

$m_{water}{C}_{pwater}(T_{water}-T_{eq})$is the heat lost by water in reaching the equilibrium temperature.

Also,

$m_{ice}{C}_{pice}(0°C-T_{ice})+\left(m_{ice}\Delta H_{fus}\right)+m_{icewater}{C}_{pwater}(T_{eq}-0°C)=m_{water}{C}_{pwater}(T_{water}-T_{eq})$

Substituting the known values,

$2mol\times \frac{38J}{Kmol}(0°C-(-10°C\left)\right)+2mol\times \frac{6007J}{mol}+2mol\times \frac{75J}{molK}\times (T_{eq}-0°C)=20mol\times \frac{75J}{mol.K}\times (20°C-T_{eq})$

On further solving,

Therefore;

$$\begin{gathered} Teq=\frac{17226J}{{\displaystyle \frac{1650J}{C}}} \\ =10.{44}^{\circ }C \end{gathered}$$

After reaching equilibrium, the temperature of the system will be $10.{44}^{\circ }C$.