A 2 kg copper block is heated to 500° C and then it is placed on a large block of ice at 0° C. 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 - kg -1, the amount of ice that can melt is

Question

A 2 kg copper block is heated to 500° C and then it is placed on a large block of ice at 0° C. 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 - kg -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} 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

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

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

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

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

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

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

87​kg

57​kg

78​kg

75​kg

Heat emitted by copper = Heat gained by ice mcΔθ=m′L ⇒m′=LmcΔθ​ Given, m=2kg,c=400J−kg−1C−1, Δθ=500,L=3.5×105J−kg−1 ∴m′=3.5×1052×400×500​=78​kg

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

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

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

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

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