To determine how much heat is released when 50.0 grams of ethanol cools from [tex]\( 90^{\circ} \text{C} \)[/tex] to [tex]\( 79^{\circ} \text{C} \)[/tex], we follow these steps:
1. Identify the given data:
- Mass of ethanol, [tex]\( m = 50.0 \)[/tex] grams.
- Specific heat capacity of liquid ethanol, [tex]\( C = 1.0 \)[/tex] J/(g·°C).
- Initial temperature, [tex]\( T_{\text{initial}} = 90^{\circ} \text{C} \)[/tex].
- Final temperature, [tex]\( T_{\text{final}} = 79^{\circ} \text{C} \)[/tex].
2. Calculate the change in temperature:
The change in temperature, [tex]\( \Delta T \)[/tex], is given by:
[tex]\[
\Delta T = T_{\text{initial}} - T_{\text{final}}
\][/tex]
Substituting the values gives:
[tex]\[
\Delta T = 90^{\circ} \text{C} - 79^{\circ} \text{C} = 11^{\circ} \text{C}
\][/tex]
3. Use the formula for heat released:
The formula for the amount of heat ([tex]\( Q \)[/tex]) released or absorbed is:
[tex]\[
Q = m \times C \times \Delta T
\][/tex]
Substituting the known values:
[tex]\[
Q = 50.0 \text{ g} \times 1.0 \text{ J/(g·°C)} \times 11^{\circ} \text{C}
\][/tex]
4. Perform the multiplication:
[tex]\[
Q = 50.0 \times 1.0 \times 11 = 550 \text{ J}
\][/tex]
Therefore, the heat released when 50.0 grams of ethanol cools from [tex]\( 90^{\circ} \text{C} \)[/tex] to [tex]\( 79^{\circ} \text{C} \)[/tex] is [tex]\( 550 \text{ J} \)[/tex].
So, the correct answer is:
[tex]\[ 550 \text{ J} \][/tex]
