A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process `(J=4.18 "joules"//cal)`

Options

A. 654 joule
B. 156.5 joule
C. `-300` joule
D. `-528.2` joule

Explanation

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. **Convert Heat from Calories to Joules**:
Given that the system receives \(300\) calories of heat, we need to convert this to joules using the conversion factor \(1 \text{ calorie} = 4.18 \text{ joules}\).

\[ \Delta Q = 300 \text{ calories} \times 4.18 \text{ joules/calorie} = 1254 \text{ joules} \]

2. **Identify Work Done**:
The work done by the system is given as \(600\) joules.

\[ W = 600 \text{ joules} \]

3. **Apply the First Law of Thermodynamics**:
Rearranging the first law equation to solve for \(\Delta U\):

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

4. **Substitute the Values**:
Now, substitute the values we calculated into the equation:

\[ \Delta U = 1254 \text{ joules} - 600 \text{ joules} \]

5. **Calculate the Change in Internal Energy**:
Performing the subtraction gives:

\[ \Delta U = 654 \text{ joules} \]

### Final Answer:
The change in the internal energy of the system is \(654\) joules.

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