If a systems internal energy increases by 250 KJ after the addition of 375 KJ of energy as heat, what was the value of thd work in process
The Correct Answer and Explanation is:
Correct Answer:
The value of the work in the process is 125 kJ.
Explanation:
To solve this, we apply the First Law of Thermodynamics, which is:
ΔU = Q – W
Where:
- ΔU is the change in internal energy of the system
- Q is the heat added to the system
- W is the work done by the system
This equation describes how energy is conserved in a thermodynamic process. When heat is added to a system, some of that energy can increase the system’s internal energy, and the rest can be used to do work on the surroundings.
From the problem:
- ΔU = 250 kJ (internal energy increases)
- Q = 375 kJ (energy added as heat)
We rearrange the equation to solve for work (W):
W = Q – ΔU
Now plug in the values:
W = 375 kJ – 250 kJ = 125 kJ
So, the system did 125 kJ of work on its surroundings.
Conceptual Understanding:
The system gained 375 kJ of heat energy. Not all of this energy was retained within the system; some of it was used to perform work. Since the internal energy only increased by 250 kJ, the remaining 125 kJ must have been used to push against external forces — for example, by expanding a gas and moving a piston. This is typical in thermodynamic systems where energy transfers occur both as heat and as work. The First Law ensures that the total energy is conserved, and this example demonstrates how to account for energy distribution between internal energy and work output.
