One kg of steam at 100°C is added to one kg of ice at 0°C. What is the compositi

Question

One kg of steam at 100°C is added to one kg of ice at 0°C. What is the composition and temperature of the resulting mixture? The heat of fusion for water is 79.7 kcal/kg, the heat ov vaporization is 539 kcal/kg, and the specific heats of ice and water are 0.5 and 1.0 kcal/kg°C, respectively.

Please explain how you got your answer. thanks so much.

Explanation / Answer

mass of steam m = 1 kg mass of ice M = 1 kg the heat of fusion for water is L = 79.7 Kcal/Kg the heat of vaporization L '= 539 Kcal/Kg the specific heats of ice is C = 0.5 Kcal/Kg°c Specific heat of water is c =1.0 Kcal/Kg°c heat lost by steam when it convert to water at 100 oC is Q = mL ' = 539 Kcal heat required to raise the temperature of ice from 0 oC to water at 100 oC is Q ' = ML + MC dt where dt = 100 oC So, Q ' = 79.7 Kcal + 50 Kcal = 129.7 Kcal Q > Q ' i.e., heat lost by steam when it convert to water at 100 oC is greater than heat required to raise the temperature of the ice from 0 oC to 100 oC . i.e., all the ice is converted to water at 100 oC by adding 1 kg of ice and 1 kg of steam. Therefore , temprature of the resulting mixture = 100 oC

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