The First Law of Thermodynamics - AP Physics 2

1. What is the First Law of Thermodynamics?

   Energy cannot be created or destroyed, only transformed. Also known as conservation of energy principle.

2. Identify the unit of energy in the First Law of Thermodynamics.

   Joule (J). Standard SI unit for all forms of energy.

3. Calculate work done by a system with $\triangle U = 30 \text{ J}$ and $Q = 40 \text{ J}$.

   $W = Q - \triangle U = 40 \text{ J} - 30 \text{ J} = 10 \text{ J}$. Rearranging the first law to solve for work done by system.

4. Calculate $\triangle U$ if $Q = -20 \text{ J}$ and $W = 10 \text{ J}$.

   $\triangle U = Q - W = -20 \text{ J} - 10 \text{ J} = -30 \text{ J}$. System loses heat and has work done on it.

5. What does $W$ represent in the First Law of Thermodynamics?

   Work done by the system. Energy transferred when system exerts force through displacement.

6. Identify the process when $Q = W$ and $\triangle U = 0$.

   Cyclic process. Process that returns to its initial thermodynamic state.

7. What is the internal energy change if a closed system releases 50 J of heat and absorbs 50 J of heat?

   $\triangle U = 0$. Equal heat in and out results in no net change.

8. What is the effect on $\triangle U$ if no heat is exchanged and the system does 25 J of work?

   $\triangle U = -W = -25 \text{ J}$. All work comes from internal energy since $Q = 0$.

9. What does $\triangle U$ represent in the First Law of Thermodynamics?

   Change in internal energy of the system. Represents the total energy change within the system.

10. Calculate $\triangle U$ if $W = -40 \text{ J}$ and $Q = 60 \text{ J}$.

    $\triangle U = Q - W = 60 \text{ J} - (-40 \text{ J}) = 100 \text{ J}$. Negative work means work done on system adds to internal energy.

11. What is the work done if $\triangle U = 0$ and $Q = 100 \text{ J}$?

    $W = Q = 100 \text{ J}$. In isothermal process, all heat becomes work output.

12. Calculate $\triangle U$ if $Q = 100 \text{ J}$ and $W = 60 \text{ J}$.

    $\triangle U = Q - W = 100 \text{ J} - 60 \text{ J} = 40 \text{ J}$. Direct application of the first law formula.

13. Determine the sign of $\triangle U$ if the system releases 30 J of heat and does 10 J of work.

    Negative. $\triangle U = -30 - 10 = -40$ J, which is negative.

14. What does $Q$ represent in the First Law of Thermodynamics?

    Heat added to the system. Energy transferred into or out of the system.

15. Identify the system's behavior if $\triangle U = 0$ and $Q = 0$.

    Isolated system. System with no energy exchange and no work done.

16. Find $Q$ if $\triangle U = -10 \text{ J}$ and $W = 5 \text{ J}$.

    $Q = \triangle U + W = -10 \text{ J} + 5 \text{ J} = -5 \text{ J}$. System releases heat since $Q$ is negative.

17. Determine $\triangle U$ for a system with $Q = 70 \text{ J}$ and $W = 50 \text{ J}$.

    $\triangle U = Q - W = 70 \text{ J} - 50 \text{ J} = 20 \text{ J}$. Net energy increase since heat input exceeds work output.

18. What is the work done if $Q = 20 \text{ J}$ and $\triangle U = 0 \text{ J}$?

    $W = Q = 20 \text{ J}$. Isothermal process where all heat becomes work.

19. What is the change in internal energy for an isolated system?

    $\triangle U = 0$. No energy exchange with surroundings means no change.

20. Which sign does $Q$ have when heat is absorbed by the system?

    Positive. Heat flows into the system, increasing its energy.

21. What happens to $\triangle U$ if a system absorbs 50 J of heat and does no work?

    $\triangle U = Q = 50 \text{ J}$. All absorbed heat increases internal energy since $W = 0$.

22. Which sign does $Q$ have when heat is released by the system?

    Negative. Heat flows out of the system, decreasing its energy.

23. What is the internal energy change when a system does 100 J of work and absorbs 50 J of heat?

    $\triangle U = 50 \text{ J} - 100 \text{ J} = -50 \text{ J}$. Internal energy decreases as more work is done than heat absorbed.

24. Find the heat added if $\triangle U = 20 \text{ J}$ and $W = 15 \text{ J}$.

    $Q = \triangle U + W = 20 \text{ J} + 15 \text{ J} = 35 \text{ J}$. Rearranging the first law to solve for heat input.

25. Determine $W$ if $Q = 30 \text{ J}$ and $\triangle U = -10 \text{ J}$.

    $W = Q - \triangle U = 30 \text{ J} + 10 \text{ J} = 40 \text{ J}$. Note: $\triangle U$ is negative, so $W = 30 - (-10) = 40$ J.

26. Which sign does $W$ have when work is done on the system?

    Negative. Surroundings perform work on the system, adding energy.

27. What is the value of $Q$ if $\triangle U = -10 \text{ J}$ and $W = -10 \text{ J}$?

    $Q = \triangle U + W = -10 \text{ J} - 10 \text{ J} = -20 \text{ J}$. Both terms are negative, indicating energy loss.

28. What is $\triangle U$ if $Q = 25 \text{ J}$ and $W = -25 \text{ J}$?

    $\triangle U = Q - W = 25 \text{ J} - (-25 \text{ J}) = 50 \text{ J}$. Work done on system plus heat input both increase energy.

29. State the formula for the First Law of Thermodynamics.

    $\triangle U = Q - W$. Where $U$ is internal energy, $Q$ is heat, and $W$ is work.

30. What is the heat exchanged if $\triangle U = 10 \text{ J}$ and $W = 10 \text{ J}$?

    $Q = \triangle U + W = 10 \text{ J} + 10 \text{ J} = 20 \text{ J}$. Total heat input considering both work and internal energy.

31. What is the value of $W$ if $Q = 0$ and $\triangle U = 15 \text{ J}$?

    $W = -\triangle U = -15 \text{ J}$. Work is negative since energy goes to internal energy.

32. What is the sign of $W$ if the system performs 20 J of work on the surroundings?

    Positive. System does work on surroundings, so $W$ is positive.

33. Identify the thermodynamic process with $Q = 0$ and $\triangle U = W$.

    Adiabatic process. Process with no heat exchange, so $\triangle U = -W$.

34. Find the internal energy change if $Q = 0$ and $W = -25 \text{ J}$.

    $\triangle U = -W = 25 \text{ J}$. Negative work means work is done on the system.

35. Which sign does $W$ have when work is done by the system?

    Positive. System expends energy to perform work on surroundings.

36. What is the internal energy change for a system doing 15 J of work and absorbing 15 J of heat?

    $\triangle U = Q - W = 15 \text{ J} - 15 \text{ J} = 0 \text{ J}$. Equal heat input and work output results in no change.

37. What occurs to $Q$ and $W$ in an isolated system?

    Both $Q$ and $W$ are zero. No energy exchange possible in completely isolated system.

38. What is the relationship between $Q$, $W$, and $\triangle U$ in a cyclic process?

    $\triangle U = 0$, so $Q = W$. System returns to initial state, so net heat equals net work.

39. Find the heat added if $\triangle U = 0 \text{ J}$ and $W = 50 \text{ J}$.

    $Q = W = 50 \text{ J}$. Isothermal condition where heat input equals work output.

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