An electric heater supplies heat to a system at a rate of 120 W. if system performs work at a rate of 80 `J S^(-1)`, the rate of increase in internal energy is

Options

A. `30 JS^(-1)`
B. `40 JS^(-1)`
C. `50 JS^(-1)`
D. `60 JS^(-1)`

Explanation

To solve the problem, we will use 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).

The formula can be expressed as:

\[ \Delta U = Q - W \]

Where:
- \( Q \) is the rate of heat supplied to the system (in watts, which is equivalent to joules per second),
- \( W \) is the rate of work done by the system (also in joules per second).

### Step-by-Step Solution:

1. **Identify the given values**:
- Rate of heat supplied, \( Q = 120 \, \text{W} \) (which is equivalent to \( 120 \, \text{J/s} \))
- Rate of work done, \( W = 80 \, \text{J/s} \)

2. **Apply the first law of thermodynamics**:
Using the formula,
\[ \Delta U = Q - W \]

3. **Substitute the known values into the equation**:
\[ \Delta U = 120 \, \text{J/s} - 80 \, \text{J/s} \]

4. **Perform the calculation**:
\[ \Delta U = 40 \, \text{J/s} \]

5. **Conclusion**:
The rate of increase in internal energy is \( 40 \, \text{J/s} \).

### Final Answer:
The rate of increase in internal energy is \( 40 \, \text{J/s} \).

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