11.6 Temperature change and heat capacity  (Page 4/6)

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:
   $Q_{\text{cold}}=m_{\mathrm{W}}{c}_{\mathrm{W}}\left(T_{\text{f}}-\text{20.0ºC}\right)\text{.}$
3. Note that 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:
   $\begin{array}{lll}\hfill Q_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill Q_{\text{cold}}& =& {\text{–Q}}_{\text{hot}},\\ m_{\mathrm{W}}{c}_{\mathrm{W}}\left(T_{\text{f}}-\text{20.0ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left(T_{\text{f}}-\text{150ºC.}\right)\end{array}$
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 $T_{\text{f}}$ ,
   $T_{\text{f}}=\frac{m_{\text{Al}}{c}_{\text{Al}}\left(\text{150ºC}\right)+m_{\mathrm{W}}{c}_{\mathrm{W}}\left(\text{20}\text{.0ºC}\right)}{m_{\text{Al}}{c}_{\text{Al}}+m_{\mathrm{W}}{c}_{\mathrm{W}}}\text{,}$

   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).

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.