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# Answer to Question #112272 in Mechanical Engineering for Oluwaseun Ayoola

Question #112272
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
2020-04-27T01:55:17-0400

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

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Assignment Expert
05.06.20, 15:44

Dear Nazer MUKTAR, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Nazer MUKTAR
27.04.20, 14:54

Thanks so much sir but why is the answer in term of c