To find the heat released when 18.02 grams of water cools from [tex]\(112^{\circ} \text{C}\)[/tex] to [tex]\(90^{\circ} \text{C}\)[/tex], we consider the specific heat capacity of liquid water and apply the concepts of heat transfer.
1. Identify Relevant Constants:
- Mass of water, [tex]\(m = 18.02 \text{ grams}\)[/tex]
- Initial temperature, [tex]\(T_{\text{initial}} = 112^{\circ} \text{C}\)[/tex]
- Final temperature, [tex]\(T_{\text{final}} = 90^{\circ} \text{C}\)[/tex]
- Specific heat capacity of liquid water, [tex]\(c_{\text{liquid}} = 4.184 \text{ J/g} \cdot \text{K}\)[/tex]
2. Calculate the Change in Temperature:
[tex]\[
\Delta T = T_{\text{initial}} - T_{\text{final}} = 112^{\circ} \text{C} - 90^{\circ} \text{C} = 22^{\circ} \text{C}
\][/tex]
3. Calculate the Heat Released:
Using the formula for heat transfer [tex]\(Q = mc\Delta T\)[/tex],
[tex]\[
Q = 18.02 \text{ g} \times 4.184 \text{ J/g} \cdot \text{K} \times 22 \text{ K}
\][/tex]
Calculate the amount of heat in joules:
[tex]\[
Q = 18.02 \times 4.184 \times 22 = 1659 \text{ J}
\][/tex]
4. Convert the Heat to Kilojoules:
[tex]\[
Q_{\text{cooling, kJ}} = \frac{Q}{1000} = \frac{1659}{1000} = 1.659 \text{ kJ}
\][/tex]
Since heat is being released (temperature is decreasing), the value is negative:
[tex]\[
\text{Heat released} = -1.659 \text{ kJ}
\][/tex]
Thus, to three significant figures, the heat released is:
[tex]\[
-1.659
\][/tex]
