A system is given 400 calories of heat and 1000 Joule of work is done by the system, then the change in internal energy of the system will be

Options

A. 680 Joule
B. 680 erg
C. 860 Joule
D. `– 860` Joule

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta Q = \Delta U + W \]

where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done by the system.

### Step-by-step Solution:

1. **Convert Heat from Calories to Joules**:
We are given that the system receives 400 calories of heat. We need to convert this to joules using the conversion factor \(1 \text{ calorie} = 4.2 \text{ joules}\).
\[ \Delta Q = 400 \text{ calories} \times 4.2 \text{ joules/calorie} = 1680 \text{ joules} \]

2. **Identify Work Done by the System**:
The work done by the system is given as \(W = 1000 \text{ joules}\).

3. **Apply the First Law of Thermodynamics**:
We can now substitute the values into the first law of thermodynamics equation:
\[ \Delta Q = \Delta U + W \]
Substituting the known values:
\[ 1680 \text{ joules} = \Delta U + 1000 \text{ joules} \]

4. **Solve for Change in Internal Energy (\(\Delta U\))**:
Rearranging the equation to solve for \(\Delta U\):
\[ \Delta U = 1680 \text{ joules} - 1000 \text{ joules} = 680 \text{ joules} \]

Thus, the change in internal energy of the system is:
\[ \Delta U = 680 \text{ joules} \]