Calculating required heats. You have a large supply super-cooled ice at -10.0 de

Question

Calculating required heats. You have a large supply super-cooled ice at -10.0 degree C to do informal experiments in your college dorm room. Thus: the initial temperature for all parts of the problem below is -10.0 degree C. You will also need the following information: Specific heat of ice Ci = 0.480 cal/(g degree C ); heat of fusion of water Lf = 79.9 cal/g, and specific heat of water Cw= 1.00 cal/(g degree C ). How much heat is required to convert 12.0 g of ice at -10.0 degree C to pure water at 55.0 degree C? How much heat is required to convert 12.0 g of ice at -10.0 degree C to a mixture of water and ice 0degree C assuming the mixture is 50 % water? In other words, half the 12 g piece of ice is melted.

Explanation / Answer

(a)

Heat required = 12 * ( 0.48*10 + 79.9 + 1*55)

                    = 1676.4 Calories

(b)

Heat required = 12 * (0.48 * 10) + (6*79.9)

                     = 537 Calories

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