A system absorbs `10 kJ` of heat at constant volume and its temperature rises from `27^(@)C` to `37^(@)C`. The `DeltaE` of reaction is

Options

A. `100 kJ`
B. `10 kJ`
C. `0`
D. `1 kJ`

Explanation

To solve the problem, we need to determine the change in internal energy (ΔE) of the system when it absorbs heat at constant volume. Here are the steps to find the solution:

### Step 1: Understand the relationship between heat absorbed and internal energy
At constant volume, the heat absorbed by the system (Q) is equal to the change in internal energy (ΔE). This is a fundamental principle in thermodynamics.

### Step 2: Identify the given values
From the problem, we know:
- Heat absorbed (Q) = 10 kJ
- The temperature change is from 27°C to 37°C, but this information is not needed to calculate ΔE at constant volume.

### Step 3: Apply the relationship
Since we are at constant volume, we can directly state:
\[ \Delta E = Q \]

### Step 4: Substitute the known value
Now, substituting the value of Q into the equation:
\[ \Delta E = 10 \, \text{kJ} \]

### Step 5: Final answer
Thus, the change in internal energy (ΔE) of the reaction is:
\[ \Delta E = 10 \, \text{kJ} \]

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