Let's solve this step-by-step:
1. Identify the Given Data:
- Mass of ethanol, \( m = 50.0 \) grams.
- Initial temperature, \( T_{\text{initial}} = 99^\circ \, \text{C} \).
- Final temperature, \( T_{\text{final}} = 79^\circ \, \text{C} \).
- Specific heat capacity of liquid ethanol, \( c_{\text{liquid}} = 1.0 \) J/g°C.
2. Determine the Temperature Change:
[tex]\[
\Delta T = T_{\text{initial}} - T_{\text{final}}
\][/tex]
[tex]\[
\Delta T = 99^\circ \, \text{C} - 79^\circ \, \text{C} = 20^\circ \, \text{C}
\][/tex]
3. Apply the Formula for Heat Released:
The formula to calculate heat \( Q \) released or absorbed is:
[tex]\[
Q = mc\Delta T
\][/tex]
Where:
- \( m \) = mass
- \( c \) = specific heat capacity
- \( \Delta T \) = temperature change
4. Substitute the Values into the Formula:
[tex]\[
Q = 50.0 \, \text{g} \times 1.0 \, \frac{\text{J}}{\text{g}^\circ \text{C}} \times 20^\circ \, \text{C}
\][/tex]
[tex]\[
Q = 50.0 \times 20 = 1000 \, \text{J}
\][/tex]
Therefore, the heat released when 50.0 g of ethanol cools from \( 99^\circ \, \text{C} \) to \( 79^\circ \, \text{C} \) is \( 1000 \) J.
The correct answer is [tex]\( 1000 \)[/tex] J.
