Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy. (J=4200 J/kcal)

Options

A. `1.302xx10^(5)J`
B. `2.302xx10^(5)J`
C. `3.302xx10^(5)J`
D. `4.302xx10^(5)J`

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 supplied to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done on the system.

### Step-by-Step Solution:

1. **Convert Heat from kcal to Joules**:
Given that heat supplied to the system is \(30 \, \text{kcal}\) and \(1 \, \text{kcal} = 4200 \, \text{J}\), we can convert the heat into joules:
\[ \Delta Q = 30 \, \text{kcal} \times 4200 \, \text{J/kcal} = 126000 \, \text{J} \]

2. **Identify the Work Done**:
The problem states that \(4200 \, \text{J}\) of external work is done on the system. Since work is done on the system, we take it as negative in our equation:
\[ W = -4200 \, \text{J} \]

3. **Apply the First Law of Thermodynamics**:
Substitute the values of \(\Delta Q\) and \(W\) into the first law equation:
\[ \Delta Q = \Delta U + W \]
\[ 126000 \, \text{J} = \Delta U - 4200 \, \text{J} \]

4. **Rearrange to Solve for \(\Delta U\)**:
To find \(\Delta U\), we rearrange the equation:
\[ \Delta U = 126000 \, \text{J} + 4200 \, \text{J} \]
\[ \Delta U = 130200 \, \text{J} \]

5. **Final Result**:
Thus, the change in internal energy of the system is:
\[ \Delta U = 130200 \, \text{J} \quad \text{or} \quad 1.302 \times 10^5 \, \text{J} \]