Question

100.0 g water cools from 85.0°C to 20.0°C. If the specific heat of water is 4.18 J/g°C, calculate the change in energy.

Answer 1

Answer:

-27.2 kJ

Explanation:

We can use the heat-transfer formula. Recall that:

$\displaystyle q = mC\Delta T$

Where m is the mass, C is the substance's specific heat, and ΔT is the change in temperature.

Hence substitute:

$\displaystyle \begin{aligned} q & = (100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g- ^\circ C}}\right)(20.0\text{ ^\circ C} - 85.0\text{ ^\circ C}) \\ \\ & =(100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g- ^\circ C}}\right)(-65.0\text{ ^\circ C}) \\ \\ & = -2.72\times 10^4\text{ J} = -27.2\text{ kJ}\end{aligned}$

Therefore, the cooling of the water released about 27.2 kJ of heat.