Question

a carnot heat engine whise simk is at 290K has an efficiency of 30%.By how much the temperature of the source should be increased to have its efficiency equal to 50% keeping sink temperature constant​

Answer 1

Answer:

You should remember the equation relating efficiency and sink and source temperature.

n = 1- T1/T2,

where n = efficiency, T1= sink temperature, T2 = source temperature.

1st case: Let the source temperature here be "x"

n = 1- T1/T2

=> 30/100 = 1- 290/x

=> 3/10 = 1- 290/x

=> 290/x = 1- 3/10 = 7/10

=> x = (290*10)/7

2nd case: Let source temperature be "y" in this case.

n = 1- T1/T2

=> 50/100 = 1- 290/y

=> 1/2 = 1- 290/y

=> 290/y = 1/2

=> y = (290*2)

We have to find (y - x):

y - x

=> (290*2) - (290*10)/7

=> 580 - (2900/7)

=> 580 - 414.28

=> 165.7 K

So the answer is 165.7 K.