Answer to Question #112196 in Mechanical Engineering for Nworgu Chigozie

Question #112196
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.
1
Expert's answer
2020-04-28T12:26:31-0400

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=  =140000 J

10000×1×14=140000J


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