Heat is added to 1.99 kg of ice at -25 Â°C. How many kilocalories are required to change the ice to steam at 144 Â°C? If you could show me how to set this problem up it would be much appreciated! I\'ve been staring at it for a while and don\'t even know where to start. Thank you! Solution mass = m = 1.99 kg = 1990g latent heat of fusion of ice = L1 = 80 cal / g latent heat of vaporization of water = L2 = 540 cal / g Specific heat of ice = C1 = 0.5 cal /g K Specific heat capacity of water = C2 = 1 ca/g K Specific heat capacity of steam = C3 = 0.48 cal/ g K heat required = m ( C1(0 + 25) + L1 + C2(100-0) + L2 + C3(144-100) = 1990 * ( 0.5*25 + 80 + 1*100 + 540 + 0.48*47) = 1502569.4calories = 1502.57 Kilocalories .