An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of $75\ ext{J/s}$, then the rate at which internal energy increases will be:

Question

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of $75\ 	ext{J/s}$, then the rate at which internal energy increases will be:

Accepted Answer

D. 1. The **first law of thermodynamics** relates the heat supplied, the change in internal energy, and the work done by a system.
$$
Q = \Delta U + W
$$

2. Since the question gives rates (power) rather than total quantities, differentiate the equation with respect to time.
$$
\frac{dQ}{dt} = \frac{d(\Delta U)}{dt} + \frac{dW}{dt}
$$

3. Substitute the given rate of heat supply (100 W) and rate of work done (75 W):
$$
100 = \frac{d(\Delta U)}{dt} + 75
$$

4. Solve for the rate of increase of internal energy.
$$
\frac{d(\Delta U)}{dt} = 25\ 	ext{W}
$$

Hence, the answer is **D**.

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s75\ \text{J/s}75 J/s, then the rate at which internal energy increases will be:

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of $75\ \text{J/s}$ , then the rate at which internal energy increases will be: