The specific heat capacity of ice is about 0.5 cal/g°C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1-g ice cube at absolute zero (?273°C) to 1 g of boiling water. How does this number of calories compare with the number of calories required to change the same gram of 100°C boiling water to 100°C steam?

Solution 2P Introductio The heat required to change ice from absolute zero to its boiling point can be divided in three steps. First heat required to change the temperature from absolute zero to melting point. Then the latent heat required to change its phase from solid to liquid and in final step, heat required to change temperature of the water from melting point to boiling point. Solution Part I Step 1 The heat required to change temperature is given by Q = mcT Now to change the temperature of the ice from absolute zero (-273°C) to melting point (0°C) is given by Q 1 mc ice ice Here we have m = 1 g is the mass of the ice, ice= 0.5 cal/g°C is the specific heat and T = (0°C ( 273°C) = 273°C is the change in temperature of the ice, therefore we have ice Q 1 (1 g)(0.5 cal/g°C)(273°C) = 136.5 cal Step 2 The heat required to melt ice is given by Q = mL 2 Where m = 1 g is the mass of the ice, L = 80 cal/g is the latent heat of the ice. Hence the required heat is Q = (1 g)(80 cal/g) = 80 cal 2 Step 3 Now, we have to supply heat to change the temperature of the water from 0°C to 100°C . The required heat is given by Q = mc T 3 water water Here we have m = 1 g is the mass of the water, c water= 1 cal/g°C is the specific heat and T = (100°C 0°) = 100°C is the change in temperature of the water. Hence, we have water Q 2 (1 g)(1 cal/g°C)(100°C) = 100 cal