First slide

Change of state

Question

A piece of ice of mass 100 g and at temperature 0⁰C is put in 200 g of water at 25⁰C. How much ice will melt
as the temperature of water reaches 0⁰C ? (Take, latent heat of fusion of ice = 3.4 X 10⁵ Jkg^-1) [UK PMT 2015]

Moderate

A

128 g
B

185.4 g
C

92.8 g
D

61.8 g

Solution

Given, mass of ice-piece, m₁ = 100 g = 0.1 kg
Mass of water, m₂ = 200 g = 0.2 kg
When temperature of mixture is 0⁰C, then heat lost by water,
$Q = m_{2} c (Δ θ) = 0.2 \times 4200 (25 - 0)$ ( $∵$ Specific heat of water = 4200 J)
$\Rightarrow$ Q = 21000J
Now, mass of melted ice, $m' = \frac{Q}{L} = \frac{21000}{3.4 \times 10^{5}} ≃ 61.8 g$

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