Easy Tutorial
For Competitive Exams

The specific heat capacity of a metal at low temperature (T) is given as Cp(kJK–1kg–1) = 32$\left(\dfrac{\text{T}}{400}\right)^3$. A 100 gram vessel of this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is

Less than 0.028 kJ
Equal to 0.002 kJ
Greater than 0.148 kJ
Between 0.148 kJ and 0.028 kJ
Share with Friends
Privacy Copyright Contact Us