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Standard X

Physics

Phase Change

A lump of ice...

Question

A lump of ice (0.1 kg) at $- 10^{\circ}$ C is put in 0.15 kg of water at $20^{\circ}$ C. How much water will be found in the mixture when it has reached thermal equilibrium?
$(C_{i c e} = 0.5 k c a l k g^{- 1} k^{- 1}, L_{i c e} = 80 k c a l k g^{- 1})$

150 g

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31.25 g

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181.25 g

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210 g

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Solution

The correct option is C 181.25 g
The heat released by water $(20^{\circ} C \to 0^{\circ} C)$
$Q_{1} = m c Δ T = 0.15 \times 1 \times 10^{3} \times (20 - 0) = 5 k c a l$
Heat absorbed by ice $(- 10^{\circ} C \to 0^{\circ} C)$
$Q_{2} = 0.1. (0.5 \times 10^{3}) . (0 - (- 10)) = 0.5 k c a l$
Heat that remains = 3 -0.5 = 2.5 k cal
The heat that remains can melt only
$\frac{2500}{80 \times 10^{3}}$ kg of ice to water = 0.03125 kg
Total water = 150 g + 31.25 g
= 181.25 g
The equilibrium temperature will be $0^{\circ} C$

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A lump of 0.10 kg of ice at -10°C is put in 0.15 kg of water at 20°C.How much water and ice will be found in the mixture when it has reached thermal equilibrium? Specific heat capacity of ice=0.50 kcal/kg and its latent heat of melting=80 kcal/kg

Q. A lump of

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A lump of ice (0.1 kg) at −10∘C is put in 0.15 kg of water at 20∘C. How much water will be found in the mixture when it has reached thermal equilibrium? (Cice=0.5 k cal kg−1k−1, Lice=80 k cal kg−1)

A lump of ice (0.1 kg) at $- 10^{\circ}$ C is put in 0.15 kg of water at $20^{\circ}$ C. How much water will be found in the mixture when it has reached thermal equilibrium?
$(C_{i c e} = 0.5 k c a l k g^{- 1} k^{- 1}, L_{i c e} = 80 k c a l k g^{- 1})$