Question 3.9: Charles has increased his activity by doing more exercise. A...
Charles has increased his activity by doing more exercise. After a session of using weights, he has a sore arm. An ice bag is filled with 125 g of ice at 0 °C. How much heat, in kilojoules, is absorbed to melt the ice and raise the temperature of the water to body temperature, 37.0 °C?
STEP 1 State the given and needed quantities.
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 125 g of ice at 0 °C | total kilojoules to melt ice at 0 °C and to raise temperature of water to 37.0 °C | combine heat from change of state (heat of fusion) and temperature change (specific heat of water) |
STEP 2 Write a plan to convert the given quantity to the needed quantity.
Total heat = kilojoules needed to melt the ice at 0 °C and heat the water from 0 °C (freezing point) to 37.0 °C
STEP 3 Write the heat conversion factors and any metric factor.
\begin{array}{r c}\boxed{\begin{matrix} 1 g of H_{2}O(s \rightarrow l) = 334 J \\\frac{\text{334 J }}{1 g of H_{2}O } \text{ and } \frac{1 g of H_{2}O }{\text{334 J }} \end{matrix}} & \boxed{\begin{matrix} 1 \mathrm{~SH_{water}} ~ = \frac{4.184 J}{g °C} \\\frac{4.184 J}{g °C} \text{ and } \frac{g °C}{4.184 J} \end{matrix}} & \boxed{\begin{matrix} \text{1 kJ = l000 J }\\ \frac{\text{l000 J }}{\text{1 kJ }}\text{ and }\frac{\text{1 kJ}}{\text{l000 J }} \end{matrix}}\end{array}STEP 4 Set up the problem and calculate the needed quantity.
\boxed{\Delta T} = 37.0°C- 0 °C= \boxed{37.0°C}Heat needed to change ice (solid) to water (liquid) at 0 °C:
\boxed{Heat} = \underset{Three SFs }{\boxed{125 \cancel{g ice}}} \times \overset{Three SFs}{\underset{Exact}{\boxed{\frac{334 \cancel{J}}{1 \cancel{g ice}}} }} \times \overset{Exact}{\underset{Exact}{\boxed{\frac{1 KJ}{1000 \cancel{J}} }}}= \underset{Three SFs}{41.8 kJ}Heat needed to warm water (liquid) from 0 °C to water (liquid) at 37.0 °C:
\boxed{Heat} = \underset{Three SFs }{\boxed{125 \cancel{g }}} \times \underset{Three SFs }{\boxed{37.0 °\cancel{C }}} \times \overset{Exact}{\underset{Exact}{\boxed{\frac{4.184 \cancel{J}}{1 \cancel{g} °\cancel{C}}} }} \times \overset{Exact}{\underset{Exact}{\boxed{\frac{1 KJ}{1000 \cancel{J}} }}}= \underset{Three SFs}{19.4 kJ}Calculate the total heat:
Melting ice at 0 °C 41.8 kJ
Heating water (0 °C to 37.0 °C) \underline{19.4 KJ}
Total heat needed 61.2 kJ