In a process where 100 J of heat is added to a system and the internal energy increases by 40 J, how much work is done by the system?
Practice Questions
1 question
Q1
In a process where 100 J of heat is added to a system and the internal energy increases by 40 J, how much work is done by the system?
60 J
40 J
100 J
140 J
Using the first law of thermodynamics, ΔU = Q - W, we have 40 J = 100 J - W, thus W = 100 J - 40 J = 60 J.
Questions & Step-by-step Solutions
1 item
Q
Q: In a process where 100 J of heat is added to a system and the internal energy increases by 40 J, how much work is done by the system?
Solution: Using the first law of thermodynamics, ΔU = Q - W, we have 40 J = 100 J - W, thus W = 100 J - 40 J = 60 J.
Steps: 7
Step 1: Understand the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). This can be written as ΔU = Q - W.
Step 2: Identify the values given in the problem. We know that 100 J of heat is added to the system (Q = 100 J) and the internal energy increases by 40 J (ΔU = 40 J).
Step 3: Substitute the known values into the first law equation: 40 J = 100 J - W.
Step 4: Rearrange the equation to solve for W (work done by the system). This means we need to isolate W on one side of the equation.
Step 5: Rearranging gives us W = 100 J - 40 J.
Step 6: Calculate the value of W: W = 60 J.
Step 7: Conclude that the work done by the system is 60 J.
