1 kg of ice at -20°C is mixed with 1kg steam at 200°C. Then find equilibrium temperature and mixture content. Please DON'T use Joules as unit while answering the question. Converting 1 kg steam at 200°C to 1 kg water at 0°C requires : (1* 0.5*100) + ( 1*540) + (1* 1 * 100 ) k cal heat = 690 kcal heat. Converting 1 kg ice at -20°C to 1 kg water at 0°C requires : (0.5 * 1* 20 ) + (1* 80) kcal heat = 90 kcal heat. Heat released by converting 1kg steam to 1 kg water at °C will be used to provide ice the required heat. Therefore, heat remaining = 690-90 kcal=600 kcal heat Now we have 600 kcal heat and 2kg of water at 0°C. ⇒600= 2* 1 * change in temp ⇒ change in temp. = 300 °C But that is not the answer. The solution in my textbook says that the heat required to ice is less than the heat supplied by steam. But I don't see that happening. In the method I have used, we convert the substance at relatively higher temperature to the same substance at relatively lower temperature as I have converted steam to water at 0°C here, and then use Q= m*s* ΔT to find the change in temperature. I don't know why but it does not seem to work here and the answer does not seem right to me either. What is wrong in this procedure? Please don't just give solution to the question ; tell me the flaw in my procedure.