Question

A $2 \, kg$ copper block is heated to $500 \,{}^{o}C$ and then it is placed on a large block of ice at $0 \,{}^{o}C$ . If the specific heat capacity of copper is $400 J kg ^{-1 \circ} C ^{-1}$ and latent heat of fusion of water is $3.5 \, \times 10^{5} \, J \, k g^{- 1}$ , the amount of ice that can melt is

• A. $\frac{7}{8} \, kg$
• B. $\frac{7}{5 \, } \, kg$
• C. $\frac{8}{7 \, } \, kg$
• D. $\frac{5}{7} \, kg$

Answer 1

Answer: (C)

Heat emitted by copper = Heat gained by ice
$mc\Delta \theta =m^{′}L$
$\Rightarrow \, \, m^{′}=\frac{m c \Delta \theta }{L}$
Given: $m=2 kg , c=400 J kg ^{-1 \circ} C ^{-1}$ ,
$\Delta \theta =500, \, L=3.5\times 10^{5} \, J \, kg^{- 1}$
$\therefore \, \, m^{′}=\frac{2 \times 400 \times 500}{3.5 \times 10^{5}}=\frac{8}{7} \, kg$