In a process where 300 J of heat is added to a system and the internal energy increases by 100 J, how much work is done by the system?

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Q: In a process where 300 J of heat is added to a system and the internal energy increases by 100 J, how much work is done by the system?

Solution: Using the first law of thermodynamics, ΔU = Q - W. Rearranging gives W = Q - ΔU. Here, W = 300 J - 100 J = 200 J.

Steps: 6

Step 1: Identify the amount of heat added to the system, which is 300 J. This is represented as Q.
Step 2: Identify the increase in internal energy of the system, which is 100 J. This is represented as ΔU.
Step 3: Use the first law of thermodynamics formula: ΔU = Q - W, where W is the work done by the system.
Step 4: Rearrange the formula to solve for W: W = Q - ΔU.
Step 5: Substitute the values into the rearranged formula: W = 300 J - 100 J.
Step 6: Calculate the work done by the system: W = 200 J.