If $80$ gm of steam is sent into $540$ g of ice, find the resultant temperature of the mixture.

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Solution

Given: If 80gm of steam is sent into 540g of ice
To find the resultant temperature of the mixture.
Solution:
As per the given criteria,
mass of ice, $m_{i} = 540 g$
mass of the steam, $m_{s} = 80 g$
We know
Temperature of ice, $T_{i} = 0^{\circ} C$
Temperature of steam, $T_{s} = 100^{\circ}$
Latent heat of vaporisation of steam, $L_{s} = 540 c a l g^{- 1}$
Latent heat of fusion of ice, $L_{f} = 80 c a l g^{-} 1$
Specific heat of ice, $s = 0.5 c a l g^{-} 1^{\circ} C^{- 1}$ .
We know
Heat gained by ice=heat lost by steam
Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)=heat lost by steam
$m_{i} L_{f} + m_{i} s (T - T_{i}) = m_{s} L_{s} ⟹ 540 \times 80 + 540 \times 0.5 (T - (0)) = 80 \times 540 ⟹ 43200 + 270 T = 43200^{\circ} C ⟹ T = 0$
So the final temperature of the mixture is $0^{\circ} C$

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