# Carnot cycle – problems and solutions

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

__Known:__

Low temperature (T_{2}) = 400 K

High temperature (T_{1}) = 800 K

Heat input (Q_{1}) = 10,000 Joule

Wanted: Work done by Carnot engine (W)

__Solution:__

The efficiency of the Carnot engine :

Work was done by Carnot engine :

W = e Q_{1}

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

2.

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

__Known :__

Low temperature (T_{L}) = 400 K

High temperature (T_{H}) = 600 K

Heat input (Q_{1}) = 600 Joule

__Wanted: __Work was done by Carnot engine (W)

__Solution :__

The efficiency of the Carnot engine :

Work done by Carnot engine :

W = e Q_{1}

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

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

__Known :__

Low temperature (T_{L}) = 350 K

High temperature (T_{H}) = 500 K

__Wanted :__ Efficiency of Carnot engine (e)

__Solution :__

Efficiency of Carnot engine :

e = (T_{H} – T_{L}) / T_{H}

e = (500 – 350) / 500

e = 150 / 500

e = 0.3

e = 30/100 = 30 %

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 :__

Low temperature (T_{L}) = 400 K

High temperature (T_{H}) = 600 K

__Wanted :__ heat output (Q_{2})

__Solution :__

Efficiency of Carnot engine :

e = (T_{H} – T_{L}) / T_{H}

e = (600 – 400) / 600

e = 200 / 600

e = 1/3

Work done by Carnot engine :

W = e Q_{1}

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

W = (1/3)(Q_{1})

3W = Q_{1}

Heat output :

Q_{2} = Q_{1} – W

Q_{2} = 3W – W

Q_{2} = 2W

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

__Known :__

Low temperature (T_{L}) = 500 K

High temperature (T_{H}) = 800 K

Heat output (Q_{2}) = 3000 Joule

__Wanted :__ Heat input (Q_{1})

__Solution :__

Efficiency of Carnot engine :

e = (T_{H} – T_{L}) / T_{H}

e = (800 – 500) /8600

e = 300 / 800

e = 3/8

Work done by Carnot engine :

W = e Q_{1}

W = (3/8)(Q_{1})

8W/3 = Q_{1 }

Q_{2} = Q_{1} – W

Q_{2} = 8W/3 – 3W/3

Q_{2} = 5W/3

3Q_{2} = 5W

W = 3Q_{2}/5 = 3(3000)/5 = 9000/5 = 1800

Heat absorbed by engine :

Q_{1} = W + Q_{2} = 1800 + 3000 = 4800 Joule

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 (T_{H}) = 800 K , efficiency (e) = 50% = 0.5

__Wanted :__ High temperature (T_{H}) if efficiency (e) = 80% = 0.8

__Solution :__

Low temperature = 400 Kelvin

What is the high temperature (T_{H}) if efficiency (e) = 80 % ?

High temperature = 2000 Kelvin

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 (T_{H}) = 600 K , efficiency (e) = 40% = 0.4

__Wanted :__ High temperature (T_{H}) if efficiency (e) = 75% = 0.75

__Solution :__

High temperature (T_{H}) if efficiency (e) = 75 % ?

High temperature = 1440 Kelvin