Question

When $1.5\, kg$ of ice at $0^{\circ}C$ mixed with $2 \,kg$ of water at $70^{\circ}C$ in a container, the resulting temperature is $5^{\circ}C$ the heat of fusion of ice is $\left(s_{water}=4186\,J\,kg^{-1}\,K^{-1}\right)$

• A. $1.42 \times 10^{5}\,J\,kg^{-1}$
• B. $2.42 \times 10^{5}\,J\,kg^{-1}$
• C. $3.42 \times 10^{5}\,J\,kg^{-1}$
• D. $4.42 \times 10^{5}\,J\,kg^{-1}$

Answer 1

Answer: (C)

Heat lost by water $=m_{w}s_{w}\left(T_{i}-T_{f}\right)$
$=2 \times 4186 \times\left(70-5\right)$
$=544180\,J$
Heat required to melt ice $=m_{i}L_{f}=1.5 \times L_{f}$
Heat required to rise temperature of ice
$=m_{i}s_{w}\,\left(T_{f}-T_{0}\right)=1.5 \times \left(4186\right) \times\left(5-0^{\circ}\right)$
$=31395\,J$
By the principle of calorimetry
Heat lost $=$ heat gained
$544180=1.5\,L_{f}+31395$
$\therefore L_{f}=\frac{512785}{1.5}=341856.67$
$=3.42 \times 10^{5}\,J\,kg^{-1}$