A block of ice at $0^{0} C$ and mass M is kept inside isolated container of negligible heat capacity. Steam of mass m at temperature $100^{0} C$ is kept side the container. The ice melted and its temperature becomes $20^{0} C$ . The steam of mass 0.2 kg at $100^{0} C$ is again let inside the container, due to which temperature inside the container becomes $30^{0} C$ in steady state. Latent heat of ice and stem are 80 kcal/kg and 540kcal/kg respectively. Specific heat capacity of water is $S_{w} = 1 k c a l / k g -^{0} C$ . The value of m (in kg) is ……….

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answer is 1.69.

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Detailed Solution

Using conservation of energy

$\begin{array}{l} (m \times 540) + m \times 1 \times (100 - 20) = (M \times 80) + M \times 1 (20 - 0) \\ \Rightarrow M + 6.2 m \\ (0.2 \times 540) + 0.2 \times 1 \times (100 - 30) = (M + m) \times 1 \times (30 - 20) \\ \Rightarrow M + m = 12.2 \\ S o l v i n g e q u a t i o n s \\ m = 1.69 k g \\ M = 10.48 k g \end{array}$