A 2 kg copper block is heated to 500 oC and then it is placed on a large block of ice at 0 oC . If the specific heat capacity of copper is 400 J kg -1 ° C -1 and latent heat of fusion of water is 3.5 × 105 J k g- 1 , the amount of ice that can melt is

Question

A 2 kg copper block is heated to 500 oC and then it is placed on a large block of ice at 0 oC . If the specific heat capacity of copper is 400 J kg -1 ° C -1 and latent heat of fusion of water is 3.5 × 105 J k g- 1 , the amount of ice that can melt is (A) (7/8) kg (B) (7/5 ) kg (C) (8/7 ) kg (D) (5/7) kg

Tardigrade Admin · Accepted Answer

Heat emitted by copper = Heat gained by ice

m c Δ θ = m^{'} L

\Rightarrow m^{'} = \frac{m c Δ θ}{L}

Given:

m = 2 k g, c = 400 J k g^{- 1 \circ} C^{- 1}

,

Δ θ = 500, L = 3.5 \times 1 0^{5} J k g^{- 1}

∴ m^{'} = \frac{2 \times 400 \times 500}{3.5 \times 1 0 ^{5}} = \frac{8}{7} k g

Q. A $2 k g$ 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 $400 J k g^{- 1 \circ} C^{- 1}$ and latent heat of fusion of water is $3.5 \times 1 0^{5} J k g^{- 1}$ , the amount of ice that can melt is

$\frac{7}{8} k g$

$\frac{7}{5} k g$

$\frac{8}{7} k g$

$\frac{5}{7} k g$

Heat emitted by copper = Heat gained by ice
$m c Δ θ = m^{'} L$
$\Rightarrow m^{'} = \frac{m c Δ θ}{L}$
Given: $m = 2 k g, c = 400 J k g^{- 1 \circ} C^{- 1}$ ,
$Δ θ = 500, L = 3.5 \times 1 0^{5} J k g^{- 1}$
$∴ m^{'} = \frac{2 \times 400 \times 500}{3.5 \times 1 0 ^{5}} = \frac{8}{7} k g$

Q. A 2kg copper block is heated to 500oC and then it is placed on a large block of ice at 0oC . If the specific heat capacity of copper is 400Jkg−1∘C−1 and latent heat of fusion of water is 3.5×105Jkg−1 , the amount of ice that can melt is

87​kg

57​kg

78​kg

75​kg