A system suffers an increase in internal energy of 80J and at the same time has 50J of work done on it. What is the heat change of the system?

Options

A. `+130J`
B. `+30J`
C. `-130J`
D. `-30J`

Answer

To solve the problem, we will use the first law of thermodynamics, which is given by the equation:

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

Where:
- \(\Delta U\) is the change in internal energy,
- \(\Delta Q\) is the heat change of the system,
- \(W\) is the work done on the system.

### Step 1: Identify the known values
From the problem statement:
- The increase in internal energy, \(\Delta U = 80 \, \text{J}\)
- The work done on the system, \(W = +50 \, \text{J}\) (since work is done on the system, it is positive)

### Step 2: Rearrange the first law of thermodynamics
We need to find \(\Delta Q\). Rearranging the equation gives us:

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

### Step 3: Substitute the known values into the equation
Now we can substitute the known values into the equation:

\[ \Delta Q = 80 \, \text{J} - 50 \, \text{J} \]

### Step 4: Calculate \(\Delta Q\)
Now, perform the calculation:

\[ \Delta Q = 30 \, \text{J} \]

### Conclusion
The heat change of the system is \(+30 \, \text{J}\).

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