A certain process releases 64.0 kJ of heat, which is transferred to the surroundings at a constant pressure and a constant temperature of 300 K. For this process `DeltaS_("surr")` is:

Options

A. 64.0 kJ
B. `-64.0` kJ
C. `-213` J/K
D. None of these

Explanation

To find the change in entropy of the surroundings (ΔS_surr) for the given process, we can use the formula:

\[ \Delta S_{\text{surr}} = \frac{Q_{\text{surr}}}{T_{\text{surr}}} \]

Where:
- \(Q_{\text{surr}}\) is the heat absorbed by the surroundings.
- \(T_{\text{surr}}\) is the temperature of the surroundings in Kelvin.

### Step-by-Step Solution:

1. **Identify the heat released**: The process releases 64.0 kJ of heat. Since this heat is released to the surroundings, we consider it as positive for the surroundings. Therefore, we convert this to Joules:
\[ Q_{\text{surr}} = 64.0 \, \text{kJ} = 64.0 \times 10^3 \, \text{J} = 64000 \, \text{J} \]

2. **Identify the temperature of the surroundings**: The temperature of the surroundings is given as 300 K:
\[ T_{\text{surr}} = 300 \, \text{K} \]

3. **Substitute the values into the entropy formula**:
\[ \Delta S_{\text{surr}} = \frac{Q_{\text{surr}}}{T_{\text{surr}}} = \frac{64000 \, \text{J}}{300 \, \text{K}} \]

4. **Calculate ΔS_surr**:
\[ \Delta S_{\text{surr}} = \frac{64000}{300} \approx 213.33 \, \text{J/K} \]

5. **Final result**: Therefore, the change in entropy of the surroundings is:
\[ \Delta S_{\text{surr}} \approx 213.33 \, \text{J/K} \]