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A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maximum work that can be done by for each 1.0 k) of hea a reversible process for each 1.0 kJ supplied by the hot source? t supplied by the hot source. (c) How much heat is discharged into the cold sink in

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Answer :

(a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

Explanation : Given,

Temperature of hot body
T_h = 800 K

Temperature of cold body
T_c = 300 K

(a) First we have to calculate the maximum efficiency of the engine.

Formula used for efficiency of the engine.


\eta =1-(T_c)/(T_h)

Now put all the given values in this formula, we get :


\eta =1-(300K)/(800K)


\eta =0.625* 100=62.5\%

(b) Now we have to calculate the maximum work done.

Formula used :


\eta =(Q_h-Q_c)/(Q_h)=(w)/(Q_h)

where,


Q_h = heat supplied by hot source = 1 KJ


Q_c = heat supplied by hot source

w = work done = ?

Now put all the given values in this formula, we get :


\eta =(w)/(Q_h)


0.625=(w)/(1KJ)


w=0.625KJ

(c) Now we have to calculate the heat discharge into the cold sink.

Formula used :


w=Q_h-Q_c


Q_c=Q_h-w


Q_c=1-0.625


Q_c=0.375KJ

Therefore, (a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

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