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5 gm of steam at $100^{0} C$ is passed into calorimeter containing liquid. Temperature of liquid rise from $32^{0} C$ to $40^{0} C$ Then water equivalent of calorimeter and contents is

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300 gram

160 gram

375 gram

40 gram

answer is B.

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Detailed Solution

$m_{s} L_{s} + m_{s} s (t_{1} - t) = m_{w} s (t - t_{2})$

$5 (540) + 5 (1) (100 - 40)$

$= M_{W} (1) (40 - 32)$

$m_{w} = 375 g m$

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