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Question
A thermally insulated pot has 150 g ice at temperature 0°C. How much steam of 100°C has to be mixed to it, so that water of temperature 50°C will be obtained? (Given : latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution
Amount of heat required in converting 150 g ice to 0°C to water at 0°C = \[150 \times 80 = 12000 \text { cal }\]
Amount of heat energy required in heating 150 g water at 0°C to 150 g water at 50°C = \[150 \times 1 \times 50 = 7500 \text { cal }\]
Total heat energy required to convert 150 g ice at 0°C to water at 50°C = 19500 cal
Let m g be the amount of steam be mixed with water to bring the final temperature of system at 50°C.
The amount of heat released in converting m g of steam at 100°C to water at 100°C = \[m \times 540 = 540m\]
The amount of heat released in converting m g of water at 100°C to water at 50°C = \[m \times \times 1 \times 50 = 50m\]
Total heat energy released to convert m g steam at 100°C to water at 50°C = 590m cal
Using the principle of calorimetry, we have
590m = 19500
\[m = \frac{19500}{590} = 33 g\]
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