1.4 Heat transfer, specific heat, and calorimetry  (Page 6/14)

Solution

1. Use the equation for heat transfer $Q=mc\text{Δ}T$ 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:
   
   $Q_{\text{hot}}=m_{\text{A1}}{c}_{\text{A1}}\left(T_{\text{f}}-150\text{°}\text{C}\right).$
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:
   
   $Q_{\text{cold}}=m_{\text{w}}{c}_{\text{w}}\left(T_{\text{f}}-20.0\text{°}\text{C}\right).$
3. Note that $Q_{\text{hot}}<0$ and $Q_{\text{cold}}>0$ and that as stated above, they must sum to zero:
   
   $\begin{array}{ccc}\hfill Q_{\text{cold}}+Q_{\text{hot}}& =\hfill & 0\hfill \\ \hfill Q_{\text{cold}}& =\hfill & \text{−}{Q}_{\text{hot}}\hfill \\ \hfill m_{\text{w}}{c}_{\text{w}}\left(T_{\text{f}}-20.0\text{°}\text{C}\right)& =\hfill & \text{−}{m}_{\text{A1}}{c}_{\text{A1}}\left(T_{\text{f}}-150\text{°}\text{C}\right).\hfill \end{array}$
4. This a linear equation for the unknown final temperature, $T_{\text{f}}.$ Solving for $T_{\text{f}},$
   
   $T_{\text{f}}=\frac{m_{\text{A1}}{c}_{\text{A1}}\left(150\text{°}\text{C}\right)+m_{\text{w}}{c}_{\text{w}}\left(20.0\text{°}\text{C}\right)}{m_{\text{A1}}{c}_{\text{A1}}+m_{\text{w}}{c}_{\text{w}}},$
   
   and insert the numerical values:
   
   $T_{\text{f}}=\frac{\left(0.500\text{kg}\right)\left(900\text{J/kg}\text{°}\text{C}\right)\left(150\text{°}\text{C}\right)+\left(0.250\text{kg}\right)\left(4186\text{J/kg}\text{°}\text{C}\right)\left(20.0\text{°}\text{C}\right)}{\left(0.500\text{kg}\right)\left(900\text{J/kg}\text{°}\text{C}\right)+\left(0.250\text{kg}\right)\left(4186\text{J/kg}\text{°}\text{C}\right)}=59.1\text{°}\text{C}.$

Significance

Summary

• Heat and work are the two distinct methods of energy transfer.
• Heat transfer to an object when its temperature changes is often approximated well by $Q=mc\text{Δ}T,$ where m is the object’s mass and c is the specific heat of the substance.