In a process where 100 J of heat is added to a system and the internal energy in

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Question: 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?

Options:

Correct Answer: 60 J

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.

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?

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?

Correct Answer: 60 J

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.

First Law of Thermodynamics – The principle that relates the change in internal energy of a system to the heat added and the work done by the system.