Answer:
About 7.0 × 10³ J or 7.0 kJ
Explanation:
We want to determine the amount of energy needed to raise the temperature of 67 grams of water from 20°C to 45°C.
We can use the heat equation:
[tex]\displaystyle q = mC\Delta T[/tex]
Where C is the specific heat of water.
Substitute and evaluate:
[tex]\displaystyle \begin{aligned} q & = (67\text{ g})\left(\frac{4.2\text{ J}}{\text{g-$^\circ$C}}\right)\left(45^\circ \text{ C}- 20.^\circ\text{ C}\right) \\ \\ & = (67\text{ g})\left(\frac{4.2\text{ J}}{\text{g-$^\circ$C}}\right)(25^\circ \text{ C}) \\ \\ & = 7.0\times 10^3 \text{ J}\end{aligned}[/tex]
Recall that there are 1000 J in a kJ. Hence:
[tex]\displaystyle \begin{aligned} q & = 7.0\times 10^3 \text{ J} \cdot \frac{1\text{ kJ}}{1000\text{ J}} \\ \\ & = 7.0 \text{ kJ}\end{aligned}[/tex]
In conclusion, it will take about 7.0 × 10³ J or 7.0 kJ of energy to raise the temperature of 67 grams of water from 20 °C to 45 °C.
