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A metal has low-temperature specific heat capacity given by C = 32 (T/400)³ J K^−1 kg^−1. If a 100 g container made of this metal is to be cooled from 20 K down to 4 K using a special refrigerator working at room temperature (27°C), the work needed for the cooling process is

1. equal to 0.002 kJ
2. greater than 0.148 kJ
3. between 0.148 kJ and 0.028 kJ
4. less than 0.028 kJ

Correct answer: less than 0.028 kJ

Solution

Heat removed Q = m*integral C dT from 4 to 20 K is only about 0.002 J. The actual work is W = m*integral (300/T - 1)*C dT = (0.1)(32/400^3)*integral(300T^2 - T^3)dT from 4 to 20 = about 0.0377 J, which is less than 0.028 kJ. So the work is 'less than 0.028 kJ'.

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