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Chapter 12: Problem 52

Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass m and a temperature of \(94.0^{\circ} {C},\) portion \({B}\) also has a mass \(m\) but a temperature of \(78.0^{\circ} {C},\) and portion C has a mass \(m_{{C}}\) and a temperature of \(34.0^{\circ} {C}\) . What must be the mass of portion \({C}\) so that the final temperature \(T_{{f}}\) of the three-portion mixture is \(T_{t}=50.0^{\circ} {C} ?\) Express your answer in terms of \(m ;\) for example, \(m_{{C}}=2.20 {m} .\)

Short Answer

Expert verified
The mass of portion C must be 4.5 times the mass of portions A or B, i.e., \(m_C = 4.5m\).

Step by step solution

01

Understanding the System

We have three portions of liquid: A, B, and C. Portions A and B have the same mass \(m\) but different temperatures (\(94.0^{\circ} C\) for A and \(78.0^{\circ} C\) for B). Portion C has a different mass \(m_{C}\) and a temperature of \(34.0^{\circ} C\). All portions are mixed in an insulated container.
02

Using the Principle of Conservation of Energy

Since the container is insulated, the total heat exchange within the system is zero. The heat lost by portions A and B equals the heat gained by portion C. This leads to the equation: \[ m c (T_A - T_f) + m c (T_B - T_f) = m_C c (T_f - T_C) \]where \(c\) is the specific heat capacity (which is the same for each portion because they are of the same liquid).
03

Canceling and Simplifying

The specific heat \(c\) cancels out, simplifying the equation to:\[ m (T_A - T_f) + m (T_B - T_f) = m_C (T_f - T_C) \]
04

Substituting Known Values

Substitute the given temperatures into the equation:\[ m (94 - 50) + m (78 - 50) = m_C (50 - 34) \]This simplifies further to:\[ m (44) + m (28) = m_C (16) \]
05

Solve for \(m_{C}\)

Combine like terms on the left side to get:\[ 72m = 16m_C \]Now, solve for \(m_C\) by dividing both sides by 16:\[ m_C = \frac{72m}{16} = 4.5m \]
06

Conclusion

The required mass of portion C is \(4.5m\), meaning the mass of portion C should be 4.5 times the mass of portions A or B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
In any closed system, like an insulated container, the total energy remains constant. This is the principle of conservation of energy. When mixing liquids of different temperatures, energy is transferred between them to achieve thermal equilibrium.
In this context, portion A and B lose energy because they are warmer than the final desired temperature, while portion C gains energy because it is cooler. The equation representing conservation of energy for this system is:
  • Heat lost by A: \( mc(T_A - T_f) \)
  • Heat lost by B: \( mc(T_B - T_f) \)
  • Heat gained by C: \( m_C c(T_f - T_C) \)
The heat loss by portions A and B must equal the heat gain by portion C, as the container is insulated, ensuring no heat escapes. This results in the equation being balanced, reinforcing the conservation principle.
Specific Heat Capacity
The specific heat capacity, denoted as \(c\), measures how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin).
For the equation in our example, specific heat capacity for the liquid remains constant since all portions are the same liquid.
When deriving the equation for mixing the three portions, \(c\) factors out because it is common to all terms, making the problem more manageable:
  • \( m c (T_A - T_f) \)
  • \( m c (T_B - T_f) \)
  • \( m_C c (T_f - T_C) \)
This cancellation makes it easier to handle the calculations. It focuses attention on the relationship between masses and temperature changes, simplifying the equation towards our desired result.
Thermal Equilibrium
Thermal equilibrium is achieved when the temperatures of all portions in the system become equal. In the exercise, when mixing the liquid portions, heat flows from the warmer portions (A and B) to the cooler portion (C) until all parts of the liquid reach the same final temperature, \(T_f = 50.0^{\circ} C\).
  • The exchange continues until the system can no longer change its temperature. This is the state of thermal equilibrium.
  • It's a key factor in determining the final condition after mixing the liquids, ensuring no further energy transfer occurs.
Achieving thermal equilibrium in an isolated system means all the exchanged heat between portions A, B, and C results in a balanced temperature, allowing us to solve for unknowns, such as the mass of portion C.

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Most popular questions from this chapter

mmh During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} {m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} {C}\) , and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to \(92^{\circ} {C}\) . The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?

What鈥檚 your normal body temperature? It may not be 98.6 \(^{\circ} \mathrm{F}\), the often-quoted average that was determined in the nineteenth century. A more recent study has reported an average temperature of 98.2 \(^{\circ} \mathrm{F}\). What is the difference between these averages, expressed in Celsius degrees?

ssm When the temperature of a coin is raised by 75 \({C}^{\circ}\) , the coin's diameter increases by \(2.3 \times 10^{-5} {m} .\) If the original diameter of the coin is \(1.8 \times 10^{-2} {m},\) find the coefficient of linear expansion.

An \(85.0-{N}\) backpack is hung from the middle of an aluminum wire, as the drawing shows. The temperature of the wire then drops by 20.0 \({C}^{\circ} .\) Find the tension in the wire at the lower temperature. Assume that the distance between the supports does not change, and ignore any thermal stress.

mmh When resting, a person has a metabolic rate of about \(3.0 \times 10^{5}\) joules per hour. The person is submerged neck-deep into a tub containing \(1.2 \times 10^{3}\) kg of water at \(21.00^{\circ} {C}\) . If the heat from the person goes only into the water, find the water temperature after half an hour.
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