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Problem 6: What is the least amount of work that must be performed to freeze one gram of water at 00C by means of refrigerator? Take the temperature of the surrounding as 370C. How much heat is passed on to the surrounding during this process?


Comprehension: Heat of fusion is the amount of energy required to melt a given mass of a solid at the melting point of that solid. When a given mass of a liquid is solidified, the same amount of heat will be released. As an example, the heat of fusion of water is 336 J/g.

Given                    Temperature of surrounding, T1 = 370C = 273 + 37 = 310 K

                                Temperature of melting water, T2 = 00C = 273 + 0 = 273 K

                                Heat of fusion for water = 336 J/g

Required             Heat removed to the surrounding, Q1

                                Work done to freeze 1 g of water, ΔW.

Formula               ΔW = Q1 – Q2      OR        Q1 = ΔW – Q2

Strategy: Both ΔW and Q2 are not known. We first find them and then use the above formula to find Q1 (heat passed on to the surrounding). As the heat of fusion of water is 336 J/g. This means that when 1 g of water melts, 336 J of heat is removed from water. Since we deal with 1 g of water, therefore, amount of heat removed from water is 336 J = Q2.

Now, the coefficient of performance for cooling is given by,

Put the values,

However, in terms of heat, the coefficient of cooling is given by,

Put the value from equation (1) and Q2,

Put values in the given formula, ΔW = Q1 – Q2, we have,

45.54 = Q1 – 336            OR           Q1 = 45.54 + 336            OR               Q1 = 381.54 J

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