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.

New worksheetPractise online with hints

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
furnacecooling waterEQ̇H​ = ?Q̇L​Ẇ = 300 MW

A power plant delivers a net power output of 300 MW with a thermal efficiency of 25%. Find the rate of heat input Q̇H from the fuel.

Ẇ = 300 MWη = 25%
Q̇H = MW
2.Easy
hot reservoircold reservoirEQH​ = 700 kJQL​ = 455 kJW = ?

A heat engine takes in 700 kJ of heat from a hot reservoir and rejects 455 kJ to a cold reservoir in each cycle. Find the net work W it produces per cycle.

QH = 700 kJQL = 455 kJ
W = kJ
3.Medium
TH​ = 550 °CTL​ = ?EQH​ = 1300 kJQL​ = 455 kJW

A reversible (Carnot) heat engine takes in 1300 kJ of heat from a reservoir at 550 °C and rejects 455 kJ per cycle. Find the temperature TL of the cold reservoir in °C.

TH = 550 °CQH = 1300 kJQL = 455 kJ
TL = °C
4.Medium
TH​ = 700 °CTL​ = 30 °CEQH​QL​η = 48%

A heat engine receives heat from a source at 700 °C and rejects heat to the surroundings at 30 °C. Its actual thermal efficiency is 48%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 700 °CTL = 30 °Cη = 48%
ηII = %
5.Hard
TH​ = 21 °CTL​ = -10 °CHPQ̇H​ = 11 kWQ̇L​Ẇ = ?

A heat pump delivers 11 kW of heat to a heating system at 21 °C, taking heat from a source at -10 °C. Its actual COP is 50% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.

TH = 21 °CTL = -10 °CηII = 50%Q̇H = 11 kW
Ẇ = kW
6.Hard
TH​ = 450 °CT0​ = 25 °CTL​ = -20 °CERWQ̇H​ = 130 kWQ̇L​ = ?

A Carnot heat engine receives 130 kW of heat from a source at 450 °C and rejects heat to the surroundings at 25 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -20 °C and also rejects heat to the surroundings at 25 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 450 °CT0 = 25 °CTL = -20 °CQ̇H = 130 kW
Q̇L = kW

Answer key

  1. 1. Q̇H = 1200 MW
  2. 2. W = 245 kJ
  3. 3. TL = 14.95 °C
  4. 4. ηII = 69.72 %
  5. 5. Ẇ = 2.319 kW
  6. 6. Q̇L = 429.8 kW

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