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

Mass per meter square of living rrom= 50 kg and total area =200

SO, total mass to be cool= $50\times200=10000 kg$

for ambient temperature we can say that change in internal energy= 0

Now,

or minimum work done by applying first law of thermodynamics we can say that

$dQ=dU+dW$

$dW=0 +mcdT$

$dW= 10000\times c(310-294)$

dW=160000c