In a certain process, 400 cal of heat are supplied to a system and the same time 105 J of mechanical work done on the system . The increase in its internal energy is

Options

A. 20 cal
B. 303 cal
C. 404 cal
D. 425 cal

Answer

To solve the problem step by step, 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 given values
- Heat supplied to the system, \(Q = 400 \, \text{cal}\)
- Work done on the system, \(W = 105 \, \text{J}\)

### Step 2: Convert work done from Joules to calories
Since the heat is given in calories, we need to convert the work done from Joules to calories. The conversion factor is:
\[ 1 \, \text{cal} = 4.2 \, \text{J} \]
Thus, we can convert \(W\):
\[ W = \frac{105 \, \text{J}}{4.2 \, \text{J/cal}} = 25 \, \text{cal} \]

### Step 3: Apply the First Law of Thermodynamics
According to the First Law of Thermodynamics:
\[ \Delta U = Q - W \]
Substituting the values we have:
\[ \Delta U = 400 \, \text{cal} - 25 \, \text{cal} \]

### Step 4: Calculate the change in internal energy
Now, we perform the calculation:
\[ \Delta U = 400 \, \text{cal} - 25 \, \text{cal} = 375 \, \text{cal} \]

### Conclusion
The increase in internal energy of the system is:
\[ \Delta U = 375 \, \text{cal} \]