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A heat engine receives heat from a heat source at 1200°C and has a thermal efficiency of 40 percent. The heat engine does maximum work equal to 500 kJ. Determine the heat supplied to the heat engine by the heat source, the heat rejected to the heat sink, and the temperature of heat sink.

Respuesta :

Answer:

Q1 = 1250 kJ

Q2 = 750 kJ

temperature = 610.8°C

Explanation:

given data

heat source temperature = 1200°C = 1473 K

thermal efficiency η  = 40 percent  = 0.4

maximum work ( ω max )= 500 kJ

to find out

heat supplied to the heat engine by the heat source, the heat rejected to the heat sink, and the temperature of heat sink

solution

we know that efficiency is express as

η = [tex]\frac{\omega max}{Q1}[/tex]     .........1

here Q1 is heat supplied and η is efficiency and ω is work

so Q1 = [tex]\frac{500}{0.4}[/tex]

Q1 = 1250 kJ

so

ω max = Q1 - Q2

and heat rejected to heat sink Q2 is

Q2 = Q1 - ω max

Q2 = 1250 - 500

Q2 = 750 kJ

and'we also know that here

efficiency η = 1 -  [tex]\frac{T2}{T1}[/tex]

0.4 = 1 -  [tex]\frac{T2}{1473}[/tex]

T2 = 883.8 K

temperature = 610.8°C

Explanation:

we know that efficiency is express as

η =     ....1

here Q1 is heat supplied and η is efficiency and s is work

so Q1 =

Q1 = 1250 kJ

so

Smax = Q1 - Q2

and heat rejected to heat sink Q2 is

Q2 = Q1 - s max

Q2 = 1250 - 500

Q2 = 750 kJ

Know that

efficiency η = 1 -  

0.4 = 1 -  

T2 = 883.8 K

temperature = 610

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