Question 11.4: Calculate an Equilibrium Temperature Goal Solve a calorimetr...
Calculate an Equilibrium Temperature
Goal Solve a calorimetry problem involving three substances at three different temperatures.
Problem Suppose 0.400 \mathrm{~kg} of water initially at 40.0^{\circ} \mathrm{C} is poured into a 0.300-\mathrm{kg} glass beaker having a temperature of 25.0^{\circ} \mathrm{C}. A 0.500-\mathrm{kg} block of aluminum at 37.0^{\circ} \mathrm{C} is placed in the water, and the system insulated. Calculate the final equilibrium temperature of the system.
Strategy The energy transfer for the water, aluminum, and glass will be designated Q_{w}, Q_{\mathrm{al}}, and Q_{g}, respectively. The sum of these transfers must equal zero, by conservation of energy. Construct a table, assemble the three terms from the given data, and solve for the final equilibrium temperature, T.
Apply Equation 11.5
\sum Q_{k}=0 (11.5)
to the system:
\begin{aligned} & Q_{w}+Q_{\mathrm{al}}+Q_{g}=0 \\ & m_{w} c_{w}\left(T-T_{w}\right)+m_{\mathrm{al}} c_{\mathrm{al}}\left(T-T_{\mathrm{al}}\right)+m_{g} c_{g}\left(T-T_{g}\right)=0 & (1) \end{aligned}
Construct a data table:
\begin{array}{lclcc} \hline Q(\mathrm{~J}) & m(\mathrm{~kg}) & c\left(\mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\right) & T_{f} & T_{i} \\ \hline Q_{w} & 0.400 & 4190 & T & 40.0^{\circ} \mathrm{C} \\ Q_{\mathrm{al}} & 0.500 & 9.00 \times 10^{2} & T & 37.0^{\circ} \mathrm{C} \\ Q_{g} & 0.300 & 837 & T & 25.0^{\circ} \mathrm{C} \\ \hline \end{array}
Using the table, substitute into Equation 1:
\begin{aligned} & \left(1.68 \times 10^{3} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)\left(T-40.0^{\circ} \mathrm{C}\right) \\ & +\left(4.50 \times 10^{2} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)\left(T-37.0^{\circ} \mathrm{C}\right) \\ & +\left(2.51 \times 10^{2} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)\left(T-25.0^{\circ} \mathrm{C}\right)=0 \\ & \left(1.68 \times 10^{3} \mathrm{~J} /{ }^{\circ} \mathrm{C}+4.50 \times 10^{2} \mathrm{~J} /{ }^{\circ} \mathrm{C}+2.51 \times 10^{2} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right) \mathrm{T} \\ & =9.01 \times 10^{4} \mathrm{~J} \\ & T=37.9^{\circ} \mathrm{C} \end{aligned}
Remarks The answer turned out to be very close to the aluminum’s initial temperature, so it would have been impossible to guess in advance whether the aluminum would lose or gain energy. Notice the way the table was organized, mirroring the order of factors in the different terms. This kind of organization helps prevent substitution errors, which are common in these problems.