$10 g r a m s$ of steam at $100^{o} C$ is mixed with $50 g m$ of ice at $0^{o} C$ then final temperature is

20^{o} C

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50^{o} C

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40^{o} C

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100^{o} C

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Solution

The correct option is C $40^{o} C$
Given,
$L a t e n t h e a t o f i c e L_{i c e} = 336 k J / k g$
$L a t e n t h e a t o f s t e a m L_{s t e a m} = 2260 k J / k g$
$S p e c i f i c h e a t o f w a t e r S_{w a t e r} = 4.18 k J / k g^{o} C$
$M a s s o f s t e a m, M_{s t e a m} = 0.01 k g$
$M a s s o f i c e, M_{i c e} = 0.05 k g$
If final temperature is $T_{f}$
Heat Loos from steam = heat gain by ice
$M_{s t e a m} L_{s t e a m} + M_{s t e a m} S_{w a t e r} (100 - T_{f}) = M_{i c e} L_{i c e} + M_{i c e} S_{w a t e r} (T_{f} - 0)$
$T_{f} = \frac{M_{s t e a m} L_{s t e a m} - M_{i c e} L_{i c e} + M_{s t e a m} S_{w a t e r} \times 100}{(M_{i c e} + M_{s t e a m}) S_{w a t e r}}$
$T_{f} = \frac{0.01 \times 2260 - 0.05 \times 336 + 0.01 \times 4.2 \times 100}{(0.01 + 0.05) \times 4.2}$
$T_{f} = {39.6}^{o} C ≅ 40^{o} C$

Final temperature of mixture is $40^{o} C$

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