An electric heater supplies heat to a system at a rate of 120 W. if system performs work at a rate of 80J `S^(-1)`, the rate of increase in internal energy is a) 30 J/S b) 40 J/S c) 50 J/S d) 60J/S

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:

\[ Q = \Delta U + W \]

Where:
- \( Q \) is the heat added to the system (in Joules per second or Watts),
- \( \Delta U \) is the change in internal energy (in Joules per second),
- \( W \) is the work done by the system (in Joules per second).

### Step-by-Step Solution:

1. **Identify the given values**:
- Heat supplied to the system, \( Q = 120 \, \text{W} \) (which is equivalent to \( 120 \, \text{J/s} \)).
- Work done by the system, \( W = 80 \, \text{J/s} \).

2. **Rearranging the first law of thermodynamics**:
We need to find the rate of increase in internal energy, \( \Delta U \). Rearranging the equation gives us:
\[ \Delta U = Q - W \]

3. **Substituting the values into the equation**:
Now, substitute the values of \( Q \) and \( W \) into the equation:
\[ \Delta U = 120 \, \text{J/s} - 80 \, \text{J/s} \]

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

5. **Conclusion**:
The rate of increase in internal energy is \( 40 \, \text{J/s} \). Therefore, the correct answer is:
\[ \text{Option B: } 40 \, \text{J/s} \]