An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
The internal energy is increasing at a rate of 25 W (or 25 J/s).
According to the First Law of Thermodynamics, the heat supplied to a system is used partly to increase its internal energy and partly to do work. The relation is given by:
ΔQ = ΔU + ΔW
where ΔQ is the heat supplied, ΔU is the change in internal energy, and ΔW is the work done by the system.
Given:
- Rate of heat supplied = 100 W = 100 J/s
- Rate of work done by system = 75 J/s
Rearranging the equation: ΔU = ΔQ - ΔW
Substituting the values: Rate of increase in internal energy = 100 - 75 = 25 J/s = 25 W
Thus, the internal energy of the system is increasing at a rate of 25 joules per second.
Explanation
This question applies the First Law of Thermodynamics (Eq. 11.3 from the textbook: ΔQ = ΔU + PΔV, which is a specific form of ΔQ = ΔU + ΔW). The key insight is that when heat is supplied to a system, part of it increases the internal energy and part is used by the system to perform work. Since the problem gives rates (power), we work with the rate form of the equation. The heater supplies energy at 100 J/s, and the system expends 75 J/s doing work, so the remaining 25 J/s must be stored as internal energy.
Solution Steps
Step 1: Write the First Law of Thermodynamics: ΔQ = ΔU + ΔW
Step 2: Rearrange to find internal energy change: ΔU = ΔQ - ΔW
Step 3: Substitute the given values: Rate = 100 J/s - 75 J/s = 25 J/s
Step 4: Express the answer: 25 J/s = 25 W