To find out how much heat is released when 60.0 grams of ethanol cools from 70°C to 43°C, we can use the heat transfer formula:
[tex]\[ Q = mc\Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat released or absorbed,
- [tex]\( m \)[/tex] is the mass of the substance (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity of the substance (in [tex]\( J/g^\circ C \)[/tex]),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in [tex]\( ^\circ C \)[/tex]).
Given the values:
- Mass of ethanol, [tex]\( m = 60.0 \)[/tex] grams,
- Specific heat capacity of liquid ethanol, [tex]\( c = 1.0 J/g^\circ C \)[/tex],
- Initial temperature, [tex]\( T_{\text{initial}} = 70.0^\circ C \)[/tex],
- Final temperature, [tex]\( T_{\text{final}} = 43.0^\circ C \)[/tex].
First, calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_{\text{initial}} - T_{\text{final}} \][/tex]
[tex]\[ \Delta T = 70.0^\circ C - 43.0^\circ C \][/tex]
[tex]\[ \Delta T = 27.0^\circ C \][/tex]
Next, use the heat transfer formula:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = 60.0 \, \text{g} \times 1.0 \, \text{J/g}^\circ \text{C} \times 27.0^\circ \text{C} \][/tex]
[tex]\[ Q = 60.0 \times 27.0 \][/tex]
[tex]\[ Q = 1620.0 \, \text{J} \][/tex]
Therefore, the amount of heat released when 60.0 grams of ethanol cools from 70°C to 43°C is [tex]\( 1620.0 \, \text{J} \)[/tex].
From the given options, the closest value to this result is 1600 J. Thus, the correct answer is:
[tex]\[ \boxed{1600 \, \text{J}} \][/tex]
