Solved Problem on Heat

Three liquids A, B, and C are at 10 °C, 24 °C, and 40 °C, respectively. It is known that:
a) Mixing equal masses of A and B, the final temperature is 14 °C;
b) Mixing masses of A and C in the ratio of m_A:m_C = 2:3, the final temperature is 30 °C.
Calculate what will be the equilibrium temperature of the mixture of B and C in the ratio of m_B:m_C = 1:2.

Problem data:

Initial temperature of liquid A; t_A = 10 °C;
Initial temperature of liquid B; t_B = 24 °C;
Initial temperature of liquid C; t_C = 40 °C;
Mixture (a):
- Mass of liquid A: m_A = m;
- Mass of liquid B: m_B = m;
- Equilibrium temperature between A AND B: t_AB = 14 °C;
Mixture (b):
- ratio between the masses of liquids A and C: \( m_{A}=\frac{2}{3}m_{C} \);
- Equilibrium temperature between A and C: t_AC = 30 °C.

Solution

When liquids are mixed, the colder liquid gains heat and increases the temperature and the hotter liquid loses heat to the colder one, the temperature decreases until they both reach the same temperature (equilibrium temperature), and the equation of heat transfer is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\left(t_{eq}-t_{0}\right)} \tag{I} \end{gather} \]

where c is the specific heat of each liquid. In the mixture (a), writing the expression (I) for each of the liquids

\[ \begin{gather} Q_{A}=m_{A}c_{A}\left(t_{AB}-t_{A}\right)\\ Q_{A}=mc_{A}\left(14-10\right)\\ Q_{A}=4mc_{A} \tag{II} \end{gather} \]

\[ \begin{gather} Q_{B}=m_{B}c_{B}\left(t_{AB}-t_{B}\right)\\ Q_{B}=mc_{B}\left(14-24\right)\\ Q_{B}=-10mc_{B} \tag{III} \end{gather} \]

Considering heat lost by the system to the environment negligible, the system is insulated, and there is only heat transfer between the liquids, as heat is energy transferred, we can use the Conservation of Energy, the sum of the heat transferred is equal to zero in a thermally insulated system, with expressions (II) and (III)

\[ \begin{gather} \sum Q=0\\[5pt] Q_{A}+Q_{B}=0\\[5pt] 4mc_{A}+\left(-10mc_{B}\right)=0\\[5pt] 4mc_{A}-10mc_{B}=0\\[5pt] 4mc_{A}=10mc_{B}\\[5pt] 4mc_{A}=10mc_{B}\\[5pt] \frac{c_{A}}{c_{B}}=\frac{10\cancel{m}}{4\cancel{m}}\\[5pt] \frac{c_{A}}{c_{B}}=\frac{5}{2} \tag{IV} \end{gather} \]

In mixture (b), writing the equation of heat transfer for each of the liquids and using the mass ratio given in the problem

\[ \begin{gather} Q_{A}=m_{A}c_{A}\left(t_{AC}-t_{A}\right)\\[5pt] Q_{A}=\frac{2}{3}m_{C}c_{A}\left(30-10\right)\\[5pt] Q_{A}=\frac{2}{3}m_{C}c_{A}.20\\[5pt] Q_{A}=\frac{40}{3}m_{C}c_{A} \tag{V} \end{gather} \]

\[ \begin{gather} Q_{C}=m_{C}c_{C}\left(t_{AC}-t_{C}\right)\\[5pt] Q_{C}=m_{C}c_{C}\left(30-40\right)\\[5pt] Q_{C}=-10m_{C}c_{C} \tag{VI} \end{gather} \]

Using Conservation of Energy again, applying to expressions (V) and (VI)

\[ \begin{gather} \sum Q=0\\[5pt] Q_{A}+Q_{C}=0\\[5pt] \frac{40}{3}m_{C}c_{A}+\left(-10m_{C}c_{C}\right)=0\\[5pt] \frac{40}{3}m_{C}c_{A}-10m_{C}c_{C}=0\\[5pt] \frac{40}{3}m_{C}c_{A}=10m_{C}c_{C}\\[5pt] \frac{c_{A}}{c_{C}}=\frac{3.10m_{C}}{40m_{C}}\\[5pt] \frac{c_{A}}{c_{C}}=\frac{30\cancel{m_{C}}}{40\cancel{m_{C}}}\\[5pt] \frac{c_{A}}{c_{C}}=\frac{3}{4} \tag{VII} \end{gather} \]

For the mixture of liquids B and C, writing the equation of heat transfer, where the ratio between the masses of the liquids is \( m_{B}=\frac{1}{2}m_{C} \)

\[ \begin{gather} Q_{B}=m_{B}c_{B}\left(t_{BC}-t_{B}\right)\\ Q_{B}=\frac{1}{2}m_{C}c_{B}\left(t_{BC}-24\;\right) \tag{VIII} \end{gather} \]

\[ \begin{gather} Q_{C}=m_{C}c_{C}\left(t_{BC}-t_{C}\right)\\ Q_{C}=m_{C}c_{C}\left(t_{BC}-40\right) \tag{IX} \end{gather} \]

Using Conservation of Energy again, applying to expressions (VIII) and (IX)

\[ \begin{gather} \sum Q=0\\ Q_{B}+Q_{C}=0\\ \frac{1}{2}m_{C}c_{B}\left(t_{BC}-24\right)+m_{C}c_{C}\left(t_{BC}-40\right)=0\\ \frac{1}{2}m_{C}c_{B}\left(t_{BC}-24\right)=-m_{C}c_{C}\left(t_{BC}-40\right)\\ \frac{1}{2}m_{C}c_{B}\left(t_{BC}-24\right)=m_{C}c_{C}\left(40-t_{BC}\right) \tag{X} \end{gather} \]

From expressions (IV) and (VII) we obtain the values of the specific heats of liquids B and C as a function of the specific heat of liquid A

\[ \frac{c_{A}}{c_{B}}=\frac{5}{2}\Rightarrow c_{B}=\frac{2}{5}c_{A} \]

\[ \frac{c_{A}}{c_{C}}=\frac{3}{4}\Rightarrow c_{C}=\frac{4}{3}c_{A} \]

substituting these values in the expression (X)

\[ \begin{gather} \frac{1}{\cancel{2}}\cancel{m_{C}}\frac{\cancel{2}}{5}\cancel{c_{A}}\left(t_{BC}-24\right)=\cancel{m_{C}}\frac{4}{3}\cancel{c_{A}}\left(40-t_{BC}\;\right)\\[5pt] \frac{1}{5}\left(t_{BC}-24\right)=\frac{4}{3}\left(40-t_{BC}\right)\\[5pt] t_{BC}-24=\frac{5.4}{3}\left(40-t_{BC}\right)\\[5pt] t_{BC}-24=\frac{20}{3}\left(40-t_{BC}\;\right)\\[5pt] t_{BC}-24=\frac{20}{3}.40-\frac{20}{3}t_{BC}\\[5pt] t_{BC}-24=\frac{800}{3}-\frac{20}{3}t_{BC}\\[5pt] t_{BC}+\frac{20}{3}t_{BC}=\frac{800}{3}+24 \end{gather} \]

factoring the value of t_BC on the left-hand side of the equation

\[ t_{BC}\left(1+\frac{20}{3}\right)=\frac{800}{3}+24 \]

on the left-hand side of the equation, multiplying and dividing the first term in parentheses by 3, and on the right-hand side of the equation multiplying and dividing the second term by 3

\[ \begin{gather} t_{BC}\left(1\times\frac{3}{3}+\frac{20}{3}\right)=\frac{800}{3}+24\times\frac{3}{3}\\ t_{BC}\left(\frac{3+20}{3}\right)=\frac{800+72}{3}\\ \frac{23}{\cancel{3}}t_{BC}=\frac{872}{\cancel{3}}\\ 23t_{BC}=872\\ t_{BC}=\frac{872}{23} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t_{BC}=37,9 °\text{C}} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .