Find the external work done by the system in kcal, when 20 kcal of heat is supplied to the system the increase in th internal energy is 8400 J (J=4200J/kcal)

Options

A. 16 kcal
B. 18 kcal
C. 20 kcal
D. 19 kcal

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. **Identify the Given Values:**
- Heat supplied to the system, \(\Delta Q = 20 \, \text{kcal}\)
- Increase in internal energy, \(\Delta U = 8400 \, \text{J}\)

2. **Convert Internal Energy from Joules to Kilo Calories:**
- We know that \(1 \, \text{kcal} = 4200 \, \text{J}\).
- To convert \(\Delta U\) to kcal:
\[ \Delta U = \frac{8400 \, \text{J}}{4200 \, \text{J/kcal}} = 2 \, \text{kcal} \]

3. **Apply the First Law of Thermodynamics:**
- Substitute the values into the equation:
\[ \Delta Q = \Delta U + W \]
\[ 20 \, \text{kcal} = 2 \, \text{kcal} + W \]

4. **Solve for Work Done (W):**
- Rearranging the equation to solve for \(W\):
\[ W = 20 \, \text{kcal} - 2 \, \text{kcal} = 18 \, \text{kcal} \]

5. **Conclusion:**
- The external work done by the system is \(W = 18 \, \text{kcal}\).

### Final Answer:
The external work done by the system is **18 kcal**.

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