Question

When $0.15\text{kg}$ of ice at $0^{\circ }\text{C}$ is mixed with $0.30\text{kg}$ of water at ${50}^{\circ }\text{C}$ in a container, the resulting temperature of the mixture is ${6.7}^{\circ }\text{C}$. Calculate the latent heat of fusion of ice. $(S_{water}=4186\text{J}{\text{kg}}^{–1}{\text{K}}^{-1})$

• A. $3.34\times {10}^5\text{J}{\text{kg}}^{–1}$
• B. $1.5\times {10}^5\text{J}{\text{kg}}^{–1}$
• C. $2\times {10}^5\text{J}{\text{kg}}^{–1}$
• D. $5.4\times {10}^5\text{J}{\text{kg}}^{–1}$

Answer 1

The correct option is A $3.34\times {10}^5\text{J}{\text{kg}}^{–1}$

Heat lost by the water:
$=m_{water}{S}_{water}(T_f–T_i)$
$=0.30\times 4186\times (50.0–6.7)$
$=54376.14\text{J}$

Heat required to melt the ice:
$=m_{ice}{L}_{fusion}=0.15L_f$

Heat required to raise the temperature of ice water to the final temperature:
$=m_{ice}{S}_{water}(T_f–T_i)$
$=0.15\times 4186\times (6.7–0)$
$=4206.93\text{J}$

As we know that, Heat lost is equal to heat gained.
$\Rightarrow 54376.14\text{J}=0.15L_f+4206.93\text{J}$
$L_f=3.34\times {10}^5\text{J}{\text{kg}}^{–1}.$