To find out how much heat is released when 40.0 grams of ethanol cools from [tex]\(-120^{\circ}C\)[/tex] to [tex]\(-136^{\circ}C\)[/tex], we follow these steps:
1. Identify the given data:
- Mass of ethanol ([tex]\(m\)[/tex]): 40.0 grams
- Initial temperature ([tex]\(T_i\)[/tex]): [tex]\(-120^{\circ}C\)[/tex]
- Final temperature ([tex]\(T_f\)[/tex]): [tex]\(-136^{\circ}C\)[/tex]
- Specific heat capacity in the solid state ([tex]\(c\)[/tex]): [tex]\(0.5 \, J/g^{\circ}C\)[/tex]
2. Calculate the temperature difference ([tex]\(\Delta T\)[/tex]):
[tex]\[
\Delta T = T_i - T_f = (-120^{\circ}C) - (-136^{\circ}C) = -120 + 136 = 16^{\circ}C
\][/tex]
3. Calculate the heat released ([tex]\(q\)[/tex]) using the formula:
[tex]\[
q = m \times c \times \Delta T
\][/tex]
Substituting the values:
[tex]\[
q = 40.0 \, g \times 0.5 \, J/g^{\circ}C \times 16^{\circ}C
\][/tex]
4. Perform the multiplication to find [tex]\(q\)[/tex]:
[tex]\[
q = 40.0 \times 0.5 \times 16 = 320.0 \, J
\][/tex]
Therefore, the amount of heat released when 40.0 grams of ethanol cools from [tex]\(-120^{\circ}C\)[/tex] to [tex]\(-136^{\circ}C\)[/tex] is 320 J.
So, the correct answer is:
[tex]\[
\boxed{320 \, J}
\][/tex]
