For a thermodynamic process `deltaQ = – 50` calorie and W = – 20 calorie. If the initial internal energy is – 30 calorie then final internal energy will be -

Options

A. 191.20 Calorie
B. – 60 Calorie
C. 100 Calorie
D. – 100 Calorie

Answer

To find the final internal energy in a thermodynamic process, we can 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.

Given:
- \(\Delta Q = -50\) calories (heat is lost),
- \(W = -20\) calories (work is done on the system),
- Initial internal energy \(U_i = -30\) calories.

We need to find the final internal energy \(U_f\).

### Step 1: Rearranging the first law of thermodynamics
We can rearrange the equation to find the change in internal energy:

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

### Step 2: Substitute the known values
Now, substitute the values of \(\Delta Q\) and \(W\):

\[ \Delta U = -50 - (-20) \]

### Step 3: Simplify the equation
This simplifies to:

\[ \Delta U = -50 + 20 = -30 \text{ calories} \]

### Step 4: Calculate the final internal energy
Now, we can find the final internal energy using the change in internal energy:

\[ U_f = U_i + \Delta U \]

Substituting the values:

\[ U_f = -30 + (-30) = -60 \text{ calories} \]

### Conclusion
The final internal energy \(U_f\) is \(-60\) calories.

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