Question

A house initially at a temperature 310k during a hot summer day must be cooled to a temperature 294k while the ambient

temperature is 310k. Obtain an expression for the minimum workdone required and determine the minimum work also

required to cool a room containing the total area of 200 meter square with equivalent mass of 50 kilogram meter square of the

living room.

Answer 1

Total mass of air required= 200\times 50=10000

200×50=10000

at ambient condition temperature will be constant so internal energy will be constant and thus from first law of thermodynamics

dQ=dU+dW

dW= mcdT

dW= $10000\times14\times1$ =140000 J

10000×1×14=140000J