# 8.2 Temperature change and heat capacity  (Page 4/6)

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Solution

1. Use the equation for heat transfer $Q=\text{mc}\text{Δ}T$ to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature:
${Q}_{\text{hot}}={m}_{\text{Al}}{c}_{\text{Al}}\left({T}_{\text{f}}-\text{150ºC}\right)\text{.}$
2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature:
${Q}_{\text{cold}}={m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)\text{.}$
3. Note that ${Q}_{\text{hot}}<0$ and ${Q}_{\text{cold}}>0$ and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water:
$\begin{array}{lll}\hfill {Q}_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill {Q}_{\text{cold}}& =& {\text{–Q}}_{\text{hot}},\\ {m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left({T}_{\text{f}}-\text{150ºC.}\right)\end{array}$
4. This an equation for the unknown final temperature, ${T}_{\text{f}}$
5. Bring all terms involving ${T}_{\text{f}}$ on the left hand side and all other terms on the right hand side. Solve for ${T}_{\text{f}}$ ,
${T}_{\text{f}}=\frac{{m}_{\text{Al}}{c}_{\text{Al}}\left(\text{150ºC}\right)+{m}_{W}{c}_{W}\left(\text{20}\text{.0ºC}\right)}{{m}_{\text{Al}}{c}_{\text{Al}}+{m}_{W}{c}_{W}}\text{,}$

and insert the numerical values:

$\begin{array}{lll}{T}_{\text{f}}& =& \frac{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{150ºC}\right)\text{+}\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)\left(\text{20.0ºC}\right)}{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)+\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)}\\ & =& \frac{\text{88430 J}}{\text{1496.5 J/ºC}}\\ & =& \text{59}\text{.1ºC.}\end{array}$

Discussion

This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to $\text{20.0ºC}$ than $\text{150ºC}$ ? The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter).

## Take-home experiment: temperature change of land and water

What heats faster, land or water?

To study differences in heat capacity:

• Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can achieve approximately equal masses by using $\text{50%}$ more water by volume.)
• Heat both (using an oven or a heat lamp) for the same amount of time.
• Record the final temperature of the two masses.
• Now bring both jars to the same temperature by heating for a longer period of time.
• Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes.

Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes.

If 25 kJ is necessary to raise the temperature of a block from $\text{25ºC}$ to $\text{30ºC}$ , how much heat is necessary to heat the block from $\text{45ºC}$ to $\text{50ºC}$ ?

The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case.

## Summary

• The transfer of heat $Q$ that leads to a change $\text{Δ}T$ in the temperature of a body with mass $m$ is $Q=\text{mc}\text{Δ}T$ , where $c$ is the specific heat of the material. This relationship can also be considered as the definition of specific heat.

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