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 (\( \Delta S_{\text{surr}} \)) when a process releases heat, we can follow these steps:

### Step 1: Identify the heat released
The problem states that a certain process releases 64.0 kJ of heat. Since this heat is released to the surroundings, we can denote this heat as:
\[ Q_{\text{surr}} = +64.0 \, \text{kJ} \]

### Step 2: Convert heat to joules
Since the standard unit of energy in thermodynamics is joules, we need to convert kilojoules to joules:
\[ Q_{\text{surr}} = 64.0 \, \text{kJ} \times 1000 \, \text{J/kJ} = 64000 \, \text{J} \]

### Step 3: Use the formula for change in entropy
The change in entropy of the surroundings can be calculated using the formula:
\[ \Delta S_{\text{surr}} = \frac{Q_{\text{surr}}}{T} \]
where \( T \) is the temperature in Kelvin. In this case, \( T = 300 \, \text{K} \).

### Step 4: Substitute the values into the formula
Now we can substitute the values of \( Q_{\text{surr}} \) and \( T \) into the formula:
\[ \Delta S_{\text{surr}} = \frac{64000 \, \text{J}}{300 \, \text{K}} \]

### Step 5: Calculate the change in entropy
Now, we perform the calculation:
\[ \Delta S_{\text{surr}} = \frac{64000}{300} \approx 213.33 \, \text{J/K} \]

### Conclusion
Thus, the change in entropy of the surroundings is approximately:
\[ \Delta S_{\text{surr}} \approx 213.33 \, \text{J/K} \]