0.7 twenty grams of ice is mixed with 100 gm of water at $30^{o} C$ . Find the equilibrium temperature of the mixture.

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Solution

Given,
Latent heat of ice, $L_{i c e} = 336 k J / k g$

Specific heat of water, $S_{w a t e r} = 4.2 k J / k g^{o} C$

Mass of water, $M_{w a t e r} = 0.1 k g$

Mass of ice, $M_{i c e} = 0.02 k g$

If final temperature is $T_{f}$

Heat Loos from steam = heat gain by ice

$M_{w a t e r} S_{w a t e r} (30 - T_{f}) = M_{i c e} L_{i c e} + M_{i c e} S_{w a t e r} (T_{f} - 0)$

$T_{f} = \frac{M_{w a t e r} S_{w a t e r} \times 30 - M_{i c e} L_{i c e}}{(M_{i c e} + M_{w a t e r}) S_{w a t e r}}$

$T_{f} = \frac{0.1 \times 4.2 \times 30 - 0.02 \times 336}{(0.1 + 0.02) \times 4.2}$

$T_{f} = {11.6}^{o} C$

Final temperature of mixture is $11.6^{o} C$

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