Carnot cycle – problems and solutions

1. If heat absorbed by the engine (Q1) = 10,000 Joule, what is the work done by the Carnot engine?

Known:Carnot cycle – problems and solutions 1

Low temperature (T2) = 400 K

High temperature (T1) = 800 K

Heat input (Q1) = 10,000 Joule

Wanted: Work done by Carnot engine (W)

Solution:

The efficiency of the Carnot engine :

Carnot cycle – problems and solutions 2

Work was done by Carnot engine :

W = e Q1

W = (1/2)(10,000) = 5000 Joule

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2.

Based on graph above, what is the work done by engine in a cycle?

Known :Carnot cycle – problems and solutions 3

Low temperature (TL) = 400 K

High temperature (TH) = 600 K

Heat input (Q1) = 600 Joule

Wanted: Work was done by Carnot engine (W)

Solution :

The efficiency of the Carnot engine :

 

Work done by Carnot engine :Carnot cycle – problems and solutions 4

W = e Q1

W = (1/3)(600) = 200 Joule

3. Based on the graph below, what is the efficiency of the Carnot engine?

Known :Carnot cycle – problems and solutions 5

Low temperature (TL) = 350 K

High temperature (TH) = 500 K

Wanted : Efficiency of Carnot engine (e)

Solution :

Efficiency of Carnot engine :

e = (TH – TL) / TH

e = (500 – 350) / 500

e = 150 / 500

e = 0.3

e = 30/100 = 30 %

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4. Based on graph below, the heat engine’s high temperature is 600 K and low temperature is 400 K. If the work done by engine is W, what is the heat output.

Known :Carnot cycle – problems and solutions 6

Low temperature (TL) = 400 K

High temperature (TH) = 600 K

Wanted : heat output (Q2)

Solution :

Efficiency of Carnot engine :

e = (TH – TL) / TH

e = (600 – 400) / 600

e = 200 / 600

e = 1/3

Work done by Carnot engine :

W = e Q1

W = work done by engine, e = efficiency, Q1 = heat input

W = (1/3)(Q1)

3W = Q1

Heat output :

Q2 = Q1 – W

Q2 = 3W – W

Q2 = 2W

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5. Based on graph below, if the heat output is 3000 Joule, what is the heat input.

Known :

Low temperature (TL) = 500 KCarnot cycle – problems and solutions 7

High temperature (TH) = 800 K

Heat output (Q2) = 3000 Joule

Wanted : Heat input (Q1)

Solution :

Efficiency of Carnot engine :

e = (TH – TL) / TH

e = (800 – 500) /8600

e = 300 / 800

e = 3/8

Work done by Carnot engine :

W = e Q1

W = (3/8)(Q1)

8W/3 = Q1

Q2 = Q1 – W

Q2 = 8W/3 – 3W/3

Q2 = 5W/3

3Q2 = 5W

W = 3Q2/5 = 3(3000)/5 = 9000/5 = 1800

Heat absorbed by engine :

Q1 = W + Q2 = 1800 + 3000 = 4800 Joule

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6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant.

Known :

If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5

Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8

Solution :

Carnot cycle – problems and solutions 8

Low temperature = 400 Kelvin

What is the high temperature (TH) if efficiency (e) = 80 % ?

Carnot cycle – problems and solutions 9

High temperature = 2000 Kelvin

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7. A Carnot engine works at high temperature 600 Kelvin with the efficiency of 40%. If the efficiency of the engine is 75% and the low temperature kept constant, what is the high temperature?

Known :

If high temperature (TH) = 600 K , efficiency (e) = 40% = 0.4

Wanted : High temperature (TH) if efficiency (e) = 75% = 0.75

Solution :

Carnot cycle – problems and solutions 10

High temperature (TH) if efficiency (e) = 75 % ?Carnot cycle – problems and solutions 11

High temperature = 1440 Kelvin