Heat engines & refrigerators worksheet with answers

Free printable Heat engines & refrigerators worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
hot reservoircold reservoirEQH​ = 2400 kJQL​ = 1680 kJW

A heat engine takes in 2400 kJ of heat from a hot reservoir and rejects 1680 kJ to a cold reservoir in each cycle. Find its thermal efficiency η.

QH = 2400 kJQL = 1680 kJ
η = %
2.Easy
houseoutside airHPQ̇H​ = 12 kWQ̇L​ = ?Ẇ

A heat pump with a coefficient of performance of 4.9 delivers 12 kW of heat to a house. Find the rate Q̇L at which it takes heat from the outside air.

COPHP = 4.9Q̇H = 12 kW
Q̇L = kW
3.Medium
TH​ = 725 °CTL​ = 30 °CEQH​ = 1200 kJQL​W = ?

A heat engine works between a hot reservoir at 725 °C and a cold reservoir at 30 °C. It takes in 1200 kJ of heat. Find the most work Wmax it could possibly produce.

TH = 725 °CTL = 30 °CQH = 1200 kJ
Wmax = kJ
4.Medium
TH​ = 800 °CTL​ = 25 °CEQH​ = 3100 kJQL​W = ?

A heat engine works between a hot reservoir at 800 °C and a cold reservoir at 25 °C. It takes in 3100 kJ of heat. Find the most work Wmax it could possibly produce.

TH = 800 °CTL = 25 °CQH = 3100 kJ
Wmax = kJ
5.Hard
TH​ = 35 °CTL​ = -5 °CHPQ̇H​ = 5.5 kWQ̇L​Ẇ = ?

A heat pump delivers 5.5 kW of heat to a heating system at 35 °C, taking heat from a source at -5 °C. Its actual COP is 30% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.

TH = 35 °CTL = -5 °CηII = 30%Q̇H = 5.5 kW
Ẇ = kW
6.Hard
TH​ = 1000 °CT0​ = 30 °CTL​ = 0 °CERWQ̇H​ = 80 kWQ̇L​ = ?

A Carnot heat engine receives 80 kW of heat from a source at 1000 °C and rejects heat to the surroundings at 30 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at 0 °C and also rejects heat to the surroundings at 30 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 1000 °CT0 = 30 °CTL = 0 °CQ̇H = 80 kW
Q̇L = kW

Answer key

  1. 1. η = 30 %
  2. 2. Q̇L = 9.551 kW
  3. 3. Wmax = 835.5 kJ
  4. 4. Wmax = 2239 kJ
  5. 5. Ẇ = 2.38 kW
  6. 6. Q̇L = 555 kW

Full step-by-step solutions: practise this topic on examproblems.com.