A metal block of mass 3.3 kg is heated to a temperature of 400°C and then block is placed on large ice block. The specific heat of metal is $0.4 J g^{- 1} K^{- 1}$ and heat of fusion of water $330 J g^{- 1}$ , the maximum amount of ice that can melt is.

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2.2 kg

1.2 kg

2 kg

1.6 kg

answer is C.

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Detailed Solution

Mass of metal block $M = 3.3 k g$
Specific heat of metal $S_{m} = 0.4 J g^{- 1} k^{- 1}$
$\begin{array}{l} = 0.4 \times 10^{3} {Jkg}^{- 1} k^{- 1} \\ = 400 {Jkg}^{- 1} k^{- 1} \end{array}$
Latent heat of fusion
$L . f = 3.3 \times 10^{5} J . k g^{- 1}$
Heat gain by ice = Heat loss by metal
$\begin{array}{l} m^{1} L f = m s_{m} Δ t \\ m^{1} = \frac{m S_{m} Δ t}{L_{f}} = 1.6 kg \end{array}$