22,320 Cal heat is supplied to 100g of ice at zero degrees centigrade. If the latent heat of fusion of ice is 80 Cal /g and the latent heat of vaporization of water are 540 Cal/g the final amount of water thus obtained and its temperature respectively are:

Explanation

Given, heat supplied = 22320 calmass of ice = 100glatent heat of fusion of ice = 80 cal/glatent heat of vaporization of water = 540 cal/gLet's first calculate the heat required to melt the ice:Heat required = mass x Latent heat of fusion = 100g x 80 cal/g = 8000 calNext, the ice has to be converted to water at 0 degree Celsius, so the heat required for this will be:Heat required = mass x specific heat x change in temperature = 100g x 1 cal/g°C x 100°C = 10000 calNow, the water at 0 degree Celsius has to be converted to water at 100 degree Celsius, so the heat required for this will be:Heat required = mass x specific heat x change in temperature = 100g x 1 cal/g°C x 100°C = 10000 calNow, all the ice has melted and we have 200g of water at 100 degree Celsius. The remaining heat will be used to convert some of this water to steam at 100 degree Celsius. The amount of water that will be converted to steam can be calculated as:Heat required to convert M grams of water to steam = M x Latent heat of vaporization = M x 540 cal/gThe remaining heat available after all the ice has melted and the water has been heated to 100 degree Celsius can be calculated as:Remaining heat = 22320 - 8000 - 10000 = 4320 calSo, we have:M x 540 = 4320M = 8gTherefore, 8g of water will be converted to steam at 100 degree Celsius. The remaining water will be at 100 degree Celsius. So, the final mixture is 92g water and 8g steam in equilibrium at 373K.