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

-27.2 kJ

Step-by-step 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* 10^4\text{ J} = -27.2\text{ kJ}\end{aligned}`

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