Chapter 3

Q. 3.9

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?


Verified Solution

STEP 1   State the given and needed quantities.

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.

1 \space g \space of \space H_{2}O \space (s \longrightarrow l) = 334 \space J     SH_{water}= \frac{4.184 \space J}{g \space °C}       1 \space kJ = 1000 \space J

\frac{334 \space J}{1 \space g \space H_{2}O}   and    \frac{1 \space g \space H_{2}O}{334 \space J}                       \frac{4.184 \space J}{g\space °C}  and  \frac{g\space °C}{4.184 \space J}      \frac{1000 \space J}{1 \space kJ}   and  \frac{1 \space kJ}{1000 \space J}

STEP 4   Set up the problem and calculate the needed quantity.
T = 37.0 °C – 0 °C = 37.0 °C
Heat needed to change ice (solid) to water (liquid) at 0 °C:

Heat = \underset{Three \space SFs}{125 \space \cancel {g \space ice}} \space \space\times \space \space\frac{\overset{Three \space SFs}{334 \space \cancel J} }{\underset{Exact}{1\space \cancel {g \space ice}} } \space \space \times \space \space \frac{\overset{Exact}{1 \space kJ} }{\underset{Exact}{1000 \space \cancel J} }\space \space =\space \space \underset{Three \space SFs}{41.8 \space kJ}

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

Heat = \underset{Three \space SFs}{125 \space \cancel {g}} \space \space\times \space \space \underset{Three \space SFs}{37.0 \space °\cancel {C}} \space \space\times \space \space\frac{\overset{Exact}{4.184 \space \cancel J} }{\underset{Exact}{\cancel g \space° \cancel {C }} } \space \space \times \space \space \frac{\overset{Exact}{1 \space kJ} }{\underset{Exact}{1000 \space \cancel J} }\space \space =\space \space \underset{Three \space SFs}{19.4 \space kJ}

Calculate the total heat:
Melting ice at 0 °C                                  41.8 kJ

Heating water (0 °C to 37.0 °C)             \frac{19.4 \space kJ}{}

Total heat needed                                    61.2 kJ

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)