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Heat Loss worksheet

Grades 6–7 · CCSS 6.EE.A.2.c

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Indoor temperature: 60 °F
- Outdoor design temperature: 11 °F
- Wall area (excluding windows and doors): 1,400 ft²
- Wall R-value: 10
- Window and door area: 150 ft²
- Window and door U-factor: 0.2
- Ceiling area (under the attic): 1,600 ft²
- Ceiling R-value: 46
- Floor area over unheated space: 0 ft²
- Floor R-value: 22
- Heated volume: 10,500 ft³
- Air changes per hour (leakage): 0.4
- Heating system efficiency: 78%
Find: Design heat loss
Answer
2.
Problem 2.
- Indoor temperature: 75 °F
- Outdoor design temperature: 10 °F
- Wall area (excluding windows and doors): 1,800 ft²
- Wall R-value: 15
- Window and door area: 220 ft²
- Window and door U-factor: 0.3
- Ceiling area (under the attic): 1,850 ft²
- Ceiling R-value: 30
- Floor area over unheated space: 0 ft²
- Floor R-value: 21
- Heated volume: 14,000 ft³
- Air changes per hour (leakage): 0.5
- Heating system efficiency: 95%
Find: Design heat loss
Answer
3.
Problem 3.
- Indoor temperature: 45 °F
- Outdoor design temperature: 9 °F
- Wall area (excluding windows and doors): 1,100 ft²
- Wall R-value: 15
- Window and door area: 180 ft²
- Window and door U-factor: 0.3
- Ceiling area (under the attic): 1,700 ft²
- Ceiling R-value: 32
- Floor area over unheated space: 0 ft²
- Floor R-value: 24
- Heated volume: 12,000 ft³
- Air changes per hour (leakage): 0.4
- Heating system efficiency: 94%
Find: Design heat loss
Answer
4.
Problem 4.
- Indoor temperature: 64 °F
- Outdoor design temperature: 9 °F
- Wall area (excluding windows and doors): 1,250 ft²
- Wall R-value: 13
- Window and door area: 140 ft²
- Window and door U-factor: 0.2
- Ceiling area (under the attic): 1,400 ft²
- Ceiling R-value: 38
- Floor area over unheated space: 0 ft²
- Floor R-value: 15
- Heated volume: 11,000 ft³
- Air changes per hour (leakage): 0.6
- Heating system efficiency: 78%
Find: Design heat loss
Answer
5.
Problem 5.
- Indoor temperature: 76 °F
- Outdoor design temperature: 12 °F
- Wall area (excluding windows and doors): 1,100 ft²
- Wall R-value: 14
- Window and door area: 200 ft²
- Window and door U-factor: 0.2
- Ceiling area (under the attic): 1,460 ft²
- Ceiling R-value: 26
- Floor area over unheated space: 0 ft²
- Floor R-value: 24
- Heated volume: 14,000 ft³
- Air changes per hour (leakage): 0.5
- Heating system efficiency: 65%
Find: Design heat loss
Answer
6.
Problem 6.
- Indoor temperature: 55 °F
- Outdoor design temperature: 12 °F
- Wall area (excluding windows and doors): 1,075 ft²
- Wall R-value: 14
- Window and door area: 180 ft²
- Window and door U-factor: 0.4
- Ceiling area (under the attic): 1,500 ft²
- Ceiling R-value: 41
- Floor area over unheated space: 0 ft²
- Floor R-value: 18
- Heated volume: 11,000 ft³
- Air changes per hour (leakage): 0.6
- Heating system efficiency: 70%
Find: Design heat loss
Answer
7.
Problem 7.
- Indoor temperature: 80 °F
- Outdoor design temperature: 10 °F
- Wall area (excluding windows and doors): 1,550 ft²
- Wall R-value: 14
- Window and door area: 240 ft²
- Window and door U-factor: 0.3
- Ceiling area (under the attic): 1,520 ft²
- Ceiling R-value: 42
- Floor area over unheated space: 0 ft²
- Floor R-value: 16
- Heated volume: 12,500 ft³
- Air changes per hour (leakage): 0.5
- Heating system efficiency: 85%
Find: Design heat loss
Answer
8.
Problem 8.
- Indoor temperature: 90 °F
- Outdoor design temperature: 10 °F
- Wall area (excluding windows and doors): 1,400 ft²
- Wall R-value: 10
- Window and door area: 200 ft²
- Window and door U-factor: 0.4
- Ceiling area (under the attic): 1,500 ft²
- Ceiling R-value: 40
- Floor area over unheated space: 0 ft²
- Floor R-value: 17
- Heated volume: 11,750 ft³
- Air changes per hour (leakage): 0.4
- Heating system efficiency: 90%
Find: Design heat loss
Answer
9.
Problem 9.
- Indoor temperature: 70 °F
- Outdoor design temperature: 7 °F
- Wall area (excluding windows and doors): 1,400 ft²
- Wall R-value: 12
- Window and door area: 200 ft²
- Window and door U-factor: 0.3
- Ceiling area (under the attic): 1,900 ft²
- Ceiling R-value: 34
- Floor area over unheated space: 0 ft²
- Floor R-value: 24
- Heated volume: 15,250 ft³
- Air changes per hour (leakage): 0.4
- Heating system efficiency: 90%
Find: Design heat loss
Answer
10.
Problem 10.
- Indoor temperature: 75 °F
- Outdoor design temperature: 11 °F
- Wall area (excluding windows and doors): 1,200 ft²
- Wall R-value: 15
- Window and door area: 160 ft²
- Window and door U-factor: 0.3
- Ceiling area (under the attic): 1,600 ft²
- Ceiling R-value: 36
- Floor area over unheated space: 0 ft²
- Floor R-value: 18
- Heated volume: 14,000 ft³
- Air changes per hour (leakage): 0.4
- Heating system efficiency: 68%
Find: Design heat loss
Answer

Answer key: Heat Loss worksheet

1. Design heat loss: 13,739 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $60 - 11$ $= 49 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,400 \times 49}{10}$ $= 6,860 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.2 \times 150 \times 49$ $= 1,470 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,600 \times 49}{46}$ $= 1,704.3 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 10,500 \times 0.4 \times 49$ $= 3,704.4 BTU/h$
6. Add up every part to get the total heat loss
  $6,860 + 1,470 + 1,704.3 + 3,704.4$ $= 13,738.7 BTU/h$
7. Divide by the 78% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{13,738.7}{0.78}$ $= 17,613.8 BTU/h$
2. Design heat loss: 24,288 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $75 - 10$ $= 65 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,800 \times 65}{15}$ $= 7,800 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.3 \times 220 \times 65$ $= 4,290 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,850 \times 65}{30}$ $= 4,008.3 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 14,000 \times 0.5 \times 65$ $= 8,190 BTU/h$
6. Add up every part to get the total heat loss
  $7,800 + 4,290 + 4,008.3 + 8,190$ $= 24,288.3 BTU/h$
7. Divide by the 95% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{24,288.3}{0.95}$ $= 25,566.7 BTU/h$
3. Design heat loss: 9,607 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $45 - 9$ $= 36 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,100 \times 36}{15}$ $= 2,640 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.3 \times 180 \times 36$ $= 1,944 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,700 \times 36}{32}$ $= 1,912.5 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 12,000 \times 0.4 \times 36$ $= 3,110.4 BTU/h$
6. Add up every part to get the total heat loss
  $2,640 + 1,944 + 1,912.5 + 3,110.4$ $= 9,606.9 BTU/h$
7. Divide by the 94% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{9,606.9}{0.94}$ $= 10,220.1 BTU/h$
4. Design heat loss: 15,389 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $64 - 9$ $= 55 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,250 \times 55}{13}$ $= 5,288.5 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.2 \times 140 \times 55$ $= 1,540 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,400 \times 55}{38}$ $= 2,026.3 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 11,000 \times 0.6 \times 55$ $= 6,534 BTU/h$
6. Add up every part to get the total heat loss
  $5,288.5 + 1,540 + 2,026.3 + 6,534$ $= 15,388.8 BTU/h$
7. Divide by the 78% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{15,388.8}{0.78}$ $= 19,729.2 BTU/h$
5. Design heat loss: 19,246 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $76 - 12$ $= 64 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,100 \times 64}{14}$ $= 5,028.6 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.2 \times 200 \times 64$ $= 2,560 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,460 \times 64}{26}$ $= 3,593.8 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 14,000 \times 0.5 \times 64$ $= 8,064 BTU/h$
6. Add up every part to get the total heat loss
  $5,028.6 + 2,560 + 3,593.8 + 8,064$ $= 19,246.4 BTU/h$
7. Divide by the 65% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{19,246.4}{0.65}$ $= 29,609.9 BTU/h$
6. Design heat loss: 13,079 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $55 - 12$ $= 43 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,075 \times 43}{14}$ $= 3,301.8 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.4 \times 180 \times 43$ $= 3,096 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,500 \times 43}{41}$ $= 1,573.2 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 11,000 \times 0.6 \times 43$ $= 5,108.4 BTU/h$
6. Add up every part to get the total heat loss
  $3,301.8 + 3,096 + 1,573.2 + 5,108.4$ $= 13,079.4 BTU/h$
7. Divide by the 70% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{13,079.4}{0.7}$ $= 18,684.8 BTU/h$
7. Design heat loss: 23,198 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $80 - 10$ $= 70 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,550 \times 70}{14}$ $= 7,750 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.3 \times 240 \times 70$ $= 5,040 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,520 \times 70}{42}$ $= 2,533.3 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 12,500 \times 0.5 \times 70$ $= 7,875 BTU/h$
6. Add up every part to get the total heat loss
  $7,750 + 5,040 + 2,533.3 + 7,875$ $= 23,198.3 BTU/h$
7. Divide by the 85% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{23,198.3}{0.85}$ $= 27,292.2 BTU/h$
8. Design heat loss: 27,368 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $90 - 10$ $= 80 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,400 \times 80}{10}$ $= 11,200 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.4 \times 200 \times 80$ $= 6,400 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,500 \times 80}{40}$ $= 3,000 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 11,750 \times 0.4 \times 80$ $= 6,768 BTU/h$
6. Add up every part to get the total heat loss
  $11,200 + 6,400 + 3,000 + 6,768$ $= 27,368 BTU/h$
7. Divide by the 90% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{27,368}{0.9}$ $= 30,408.9 BTU/h$
9. Design heat loss: 21,568 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $70 - 7$ $= 63 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,400 \times 63}{12}$ $= 7,350 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.3 \times 200 \times 63$ $= 3,780 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,900 \times 63}{34}$ $= 3,520.6 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 15,250 \times 0.4 \times 63$ $= 6,917.4 BTU/h$
6. Add up every part to get the total heat loss
  $7,350 + 3,780 + 3,520.6 + 6,917.4$ $= 21,568 BTU/h$
7. Divide by the 90% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{21,568}{0.9}$ $= 23,964.4 BTU/h$
10. Design heat loss: 17,488 BTU/h
1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster
  $75 - 11$ $= 64 °F$
2. Walls: area times the gap, divided by the R-value (more insulation, less loss)
  $\frac{1,200 \times 64}{15}$ $= 5,120 BTU/h$
3. Windows and doors: the U-factor times the area times the gap
  $0.3 \times 160 \times 64$ $= 3,072 BTU/h$
4. Ceiling: area times the gap, divided by the R-value
  $\frac{1,600 \times 64}{36}$ $= 2,844.4 BTU/h$
5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)
  $0.018 \times 14,000 \times 0.4 \times 64$ $= 6,451.2 BTU/h$
6. Add up every part to get the total heat loss
  $5,120 + 3,072 + 2,844.4 + 6,451.2$ $= 17,487.6 BTU/h$
7. Divide by the 68% efficiency to get the input rating you need, since output is input times efficiency
  $\frac{17,487.6}{0.68}$ $= 25,717.1 BTU/h$