Q143 / 170 · Manipal (MET) 2017 · Physics
A 2 kg copper block is heated to $500^{\circ} \mathrm{C}$ and then it is placed on a large block of ice at $0^{\circ} \mathrm{C}$. If the specific heat capacity of copper is water is $3.5 \times 10^5 \mathrm{~J}^{-\mathrm{kg}^{-1}}$, the amount of ice that can melt is
- A$\frac{7}{8} \mathrm{~kg}$
- B$\frac{7}{5} \mathrm{~kg}$
- Correct answerC$\frac{8}{7} \mathrm{~kg}$
- D$\frac{5}{7} \mathrm{~kg}$
Solution
Heat emitted by copper = Heat gained by ice
$m c \Delta \theta=m^{\prime} L$
$\Rightarrow m^{\prime}=\frac{m c \Delta \theta}{L}$
$\begin{aligned}
& \text { Given, } m=2 \mathrm{~kg}, c=400 \mathrm{~J}^{-\mathrm{kg}^{-1} \mathrm{C}^{-1}} \\
& \Delta \theta=500, L=3.5 \times 10^5 \mathrm{~J}-\mathrm{kg}^{-1} \\
& \therefore \quad m^{\prime}= \frac{2 \times 400 \times 500}{3.5 \times 10^5}=\frac{8}{7} \mathrm{~kg}
\end{aligned}$
Hence, the correct answer is option: 'C'.