If heat given to a system is 6 kcal and work done is 6 k J . Then change in internal energy is

Options

A. 19.1kj
B. 12.5kj
C. 25kj
D. Zero

Answer

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

\[ \Delta U = Q - W \]

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

### Step 1: Identify the values given in the problem.
- Heat given to the system, \(Q = 6 \text{ kcal}\)
- Work done by the system, \(W = 6 \text{ kJ}\)

### Step 2: Convert heat from kcal to kJ.
Since the units of work done are in kJ, we need to convert the heat from kcal to kJ. The conversion factor is:
\[ 1 \text{ kcal} = 4.2 \text{ kJ} \]
Thus,
\[ Q = 6 \text{ kcal} \times 4.2 \text{ kJ/kcal} = 25.2 \text{ kJ} \]

### Step 3: Substitute the values into the first law of thermodynamics equation.
Now that we have both \(Q\) and \(W\) in the same units (kJ), we can substitute them into the equation:
\[ \Delta U = Q - W = 25.2 \text{ kJ} - 6 \text{ kJ} \]

### Step 4: Calculate the change in internal energy.
Now, we perform the calculation:
\[ \Delta U = 25.2 \text{ kJ} - 6 \text{ kJ} = 19.2 \text{ kJ} \]

### Final Answer:
The change in internal energy, \(\Delta U\), is \(19.2 \text{ kJ}\).

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