Answer:
-27.2 kJ
Explanation:
We can use the heat-transfer formula. Recall that:
[tex]\displaystyle q = mC\Delta T[/tex]
Where m is the mass, C is the substance's specific heat, and ΔT is the change in temperature.
Hence substitute:
[tex]\displaystyle \begin{aligned} q & = (100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g-$^\circ$C}}\right)(20.0\text{ $^\circ$C} - 85.0\text{ $^\circ$C}) \\ \\ & =(100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g-$^\circ$C}}\right)(-65.0\text{ $^\circ$C}) \\ \\ & = -2.72\times 10^4\text{ J} = -27.2\text{ kJ}\end{aligned}[/tex]
Therefore, the cooling of the water released about 27.2 kJ of heat.
