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Q.

1 g of steam at 100^oC is passed into a calorimeter of water equivalent 10g containing water of mass 90 g. Find the resultant temperature of the mixture if the initial temperature of water is 27^oC.

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a

38.4^oC

b

39.4^oC

c

33.7^oC

d

36.2^oC

answer is D.

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Detailed Solution

$\begin{array}{l} M_{S} \cdot L_{S} + M_{S} \cdot S (100 - t) = M_{w} S_{w} \\ (t - 27) + M_{c} S_{c} (t - 27) ∴ t = 33 {.74}^{\circ} c \end{array}$

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