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?

Question

Accepted Answer

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM	Given	Need	Connect
ANALYZE THE PROBLEM	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.

[latex]\begin{array}{r c}\boxed{\begin{matrix} 1 g of H_{2}O(s ightarrow l) = 334 J \\frac{ ext{334 J }}{1 g of H_{2}O } ext{ and } \frac{1 g of H_{2}O }{ ext{334 J }} \end{matrix}} & \boxed{\begin{matrix} 1 \mathrm{~SH_{water}} ~ = \frac{4.184 J}{g °C} \\frac{4.184 J}{g °C} ext{ and } \frac{g °C}{4.184 J} \end{matrix}} & \boxed{\begin{matrix} ext{1 kJ = l000 J }\ \frac{ ext{l000 J }}{ ext{1 kJ }} ext{ and }\frac{ ext{1 kJ}}{ ext{l000 J }} \end{matrix}}\end{array} [/latex]

STEP 4 Set up the problem and calculate the needed quantity.

[latex] \boxed{\Delta T} = 37.0°C- 0 °C= \boxed{37.0°C} [/latex]

Heat needed to change ice (solid) to water (liquid) at 0 °C:

[latex] \boxed{Heat} = \underset{Three SFs }{\boxed{125 \cancel{g ice}}} imes \overset{Three SFs}{\underset{Exact}{\boxed{\frac{334 \cancel{J}}{1 \cancel{g ice}}} }} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{1 KJ}{1000 \cancel{J}} }}}= \underset{Three SFs}{41.8 kJ} [/latex]

Heat needed to warm water (liquid) from 0 °C to water (liquid) at 37.0 °C:

[latex] \boxed{Heat} = \underset{Three SFs }{\boxed{125 \cancel{g }}} imes \underset{Three SFs }{\boxed{37.0 °\cancel{C }}} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{4.184 \cancel{J}}{1 \cancel{g} °\cancel{C}}} }} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{1 KJ}{1000 \cancel{J}} }}}= \underset{Three SFs}{19.4 kJ} [/latex]

Calculate the total heat:

Melting ice at 0 °C 41.8 kJ

Heating water (0 °C to 37.0 °C) [latex] \underline{19.4 KJ}[/latex]

Total heat needed 61.2 kJ

Question 3.9: Charles has increased his activity by doing more exercise. A...

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