30 The first law of thermodynamics – problems and solutions
1. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system?
Known :
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule
Wanted: the change in internal energy of the system
Solution :
The equation of the first law of thermodynamics
ΔU = Q-W
The sign conventions :
Q is positive if the heat added to the system
W is positive if work is done by the system
Q is negative if heat leaves the system
W is negative if work is done on the system
The change in internal energy of the system :
ΔU = 3000-2500
ΔU = 500 Joule
Internal energy increases by 500 Joule.
2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?
Known :
Heat (Q) = +2000 Joule
Work (W) = -2500 Joule
Wanted: The change in internal energy of the system
Solution :
ΔU = Q-W
ΔU = 2000-(-2500)
ΔU = 2000+2500
ΔU = 4500 Joule
Internal energy increases by 4500 Joule.
3. 2000 J of heat leaves the system and 2500 J of work is done on the system. What is the change in internal energy of the system?
Known :
Heat (Q) = -2000 Joule
Work (W) = -3000 Joule
Wanted: The change in internal energy of the system
Solution :
ΔU = Q-W
ΔU = -2000-(-3000)
ΔU = -2000+3000
ΔU = 1000 Joule
Internal energy increases by 4500 Joule.
Conclusion :
– If heat is added to the system, then the internal energy of the system increases
– If heat leaves the system, then the internal energy of the system decreases
– If the work is done by the system, then the internal energy of the system decreases
– If the work is done on the system, then the internal energy of the system increases
4. Calculate the change in internal energy of 2 moles of an ideal gas when 400 J of heat is added, and the gas expands, doing 300 J of work on its surroundings.
Solution:
Using the first law of thermodynamics: \( \Delta U = Q – W \):
\[ \Delta U = 400 – 300 = 100\, \text{J} \]
5. Determine the heat transfer for a system that performs 200 J of work on its surroundings and has a change in internal energy of 50 J.
Solution:
Using the first law of thermodynamics:
\[ Q = \Delta U + W = 50 + 200 = 250\, \text{J} \]
6. Calculate the work done by a system when it absorbs 600 J of heat and its internal energy increases by 150 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = 600 – 150 = 450\, \text{J}
\]
7. Determine the heat transfer for a system that performs 500 J of work and its internal energy decreases by 100 J.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = (-100) + 500 = 400\, \text{J}
\]
8. Calculate the change in internal energy when 300 J of heat is lost and 200 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = -300 + 200 = -100\, \text{J}
\]
9. Determine the work done on a system when it loses 400 J of heat and its internal energy decreases by 200 J.
Solution:
Using the first law of thermodynamics:
\[
W = \Delta U – Q = (-200) – (-400) = 200\, \text{J}
\]
10. Calculate the heat transfer when a system’s internal energy increases by 100 J and 50 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = 100 + 50 = 150\, \text{J}
\]
11. Determine the change in internal energy when a system absorbs 250 J of heat and performs 150 J of work on its surroundings.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = Q – W = 250 – 150 = 100\, \text{J}
\]
12. Calculate the work done by the system when it loses 300 J of heat, and its internal energy decreases by 100 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = -300 – (-100) = -200\, \text{J}
\]
13. Determine the heat transfer for a system that does 400 J of work on its surroundings and its internal energy increases by 150 J.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = 150 + 400 = 550\, \text{J}
\]
14. Calculate the change in internal energy when 500 J of heat is added, and the system does 300 J of work on its surroundings.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = Q – W = 500 – 300 = 200\, \text{J}
\]
15. Determine the work done by the system when it absorbs 600 J of heat, and its internal energy increases by 200 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = 600 – 200 = 400\, \text{J}
\]
16. Calculate the heat transfer when a system performs 700 J of work and its internal energy decreases by 300 J.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = (-300) + 700 = 400\, \text{J}
\]
17. Determine the change in internal energy when 800 J of heat is lost and 400 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = -800 + 400 = -400\, \text{J}
\]
18. Calculate the work done on the system when it loses 900 J of heat, and its internal energy decreases by 500 J.
Solution:
Using the first law of thermodynamics:
\[
W = \Delta U – Q = (-500) – (-900) = 400\, \text{J}
\]
19. Determine the heat transfer when a system’s internal energy increases by 600 J, and 300 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = 600 + 300 = 900\, \text{J}
\]
20. Calculate the change in internal energy when a system absorbs 700 J of heat and performs 350 J of work on its surroundings.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = Q – W = 700 – 350 = 350\, \text{J}
\]
21. Determine the work done by the system when it loses 800 J of heat, and its internal energy decreases by 400 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = -800 – (-400) = -400\, \text{J}
\]
22. Calculate the heat transfer when a system does 900 J of work on its surroundings, and its internal energy increases by 450 J.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = 450 + 900 = 1350\, \text{J}
\]
23. Determine the change in internal energy when 1000 J of heat is added, and the system does 500 J of work on its surroundings.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = Q – W = 1000 – 500 = 500\, \text{J}
\]
24. Calculate the work done by the system when it absorbs 1100 J of heat, and its internal energy increases by 550 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = 1100 – 550 = 550\, \text{J}
\]
25. Determine the heat transfer when a system performs 1200 J of work, and its internal energy decreases by 600 J.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = (-600) + 1200 = 600\, \text{J}
\]
26. Calculate the change in internal energy when 1300 J of heat is lost, and 650 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = -1300 + 650 = -650\, \text{J}
\]
27. Determine the work done on the system when it loses 1400 J of heat, and its internal energy decreases by 700 J.
Solution:
Using the first law of thermodynamics:
\[
W = \Delta U – Q = (-700) – (-1400) = 700\, \text{J}
\]
28. Calculate the heat transfer when a system’s internal energy increases by 800 J, and 400 J of work is done on the system.
Solution:
Using the first law of thermodynamics:
\[
Q = \Delta U + W = 800 + 400 = 1200\, \text{J}
\]
29. Determine the change in internal energy when a system absorbs 1500 J of heat and performs 750 J of work on its surroundings.
Solution:
Using the first law of thermodynamics:
\[
\Delta U = Q – W = 1500 – 750 = 750\, \text{J}
\]
30. Calculate the work done by the system when it loses 1600 J of heat, and its internal energy decreases by 800 J.
Solution:
Using the first law of thermodynamics:
\[
W = Q – \Delta U = -1600 – (-800) = -800\, \text{J}
\]
These problems and solutions are designed to provide a comprehensive understanding of the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system.
