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Chapter 11: Problem 64

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.

Short Answer

Expert verified
The equilibrium temperature when equal masses of the first and third liquid are mixed can be found by substituting the known values of \( c_1 \), \( c_2 \), \( T_1 \), and \( T_3 \) into the equation \( T_{13} = \frac{{c_1 T_1 + c_2 T_3}}{{c_1 + c_2}} \).

Step by step solution

01

Identifying given values

First, identify the given information from the problem. In this case, the given information includes the initial temperatures of three liquids and the equilibrium temperatures when the 1st and 2nd liquids and when the 2nd and 3rd liquids are mixed.
02

Applying the principle of caloricity

Now we'll find the specific heat capacities (also known as caloricities) of the first and second liquid, and the second and third liquid, respectively. The theory will be according to the equation \( Q = mc\Delta T \) where m is the mass, c is the specific heat capacity and 螖T is the temperature difference. In our case, as equal masses have been used, m will cancel out. So we have \( c_1 = \frac{{T_2 - T_{12}}}{{T_1 - T_{12}}} \) for the first calculation, where \( T_{12} \) is the equilibrium temperature when the first and second liquids mix, and \( c_2 = \frac{{T_3 - T_{23}}}{{T_2 - T_{23}}} \) for the second calculation, where \( T_{23} \) is the equilibrium temperature when second and third liquids mix.
03

Calculate the equilibrium temperature

To find the equilibrium temperature when the first and third liquids are mixed, \( T_{13} \), apply the caloricity principle once more: \( c_1 (T_{13} - T_1) = c_2 (T_3 - T_{13}) \). You can solve this equation for \( T_{13} \), which is the desired equilibrium temperature.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermal physics that describes the amount of heat required to raise the temperature of a given mass of a substance by one degree Celsius. This property is crucial because it determines how a substance responds to heat transfer. The formula used is:
\[ Q = mc\Delta T \]
Where:
  • \(Q\) is the heat added or removed,
  • \(m\) is the mass of the substance,
  • \(c\) is the specific heat capacity,
  • \(\Delta T\) is the change in temperature.
Specific heat capacity can vary greatly between substances. For instance, water has a high specific heat capacity, meaning it takes more energy to change its temperature. In the context of the exercise, since all liquids have equal masses, the mass \(m\) cancels out when using the formula, simplifying calculations. Here, learning to calculate "caloricity," or specific heat capacity for each liquid, helps us understand how different liquids will settle into thermal equilibrium when mixed.
Caloricity
Caloricity is another term used to describe specific heat capacity. It refers to the ability of a substance to either store or release heat or thermal energy. In the given exercise, we use caloricity to analyze how the liquids interact when mixed.
When two different substances are combined, their specific heat capacities will indicate how they share and equilibrate heat.
  • If a substance has a high caloricity, it will absorb more heat before its temperature changes significantly.
  • If a substance has a lower caloricity, it will heat up or cool down more rapidly.
In solving the problem, by setting the caloricity equations for both scenarios involving two liquids, we can deduce the specific heat capacities.
This information then allows us to predict the behavior when a new pair of liquids are mixed, aiding us in determining the equilibrium temperature.
Equilibrium Temperature
The equilibrium temperature is the final stable temperature achieved when two substances at different temperatures are mixed. It is a key outcome when considering how heat is transferred between substances until thermal equilibrium is reached.
Here are some vital points regarding equilibrium temperature:
  • It represents a balance where the heat lost by the hotter substance equals the heat gained by the cooler one.
  • In our exercise, this concept helps predict what the temperature will be when new combinations of liquids are mixed.
Take, for example, the exercise where two different sets of equal mass liquids are mixed, and two different equilibrium temperatures are provided. Understanding equilibrium temperature allows us to set up and solve equations that involve their specific heat capacities.
By solving these equations, we find the equilibrium temperature when new pairings of these liquids are combined. This problem-solving approach provides insight into the dynamic nature of heat transfer and equilibrium.

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Most popular questions from this chapter

A \(200-\mathrm{g}\) aluminum cup contains \(800 \mathrm{~g}\) of water in thermal equilibrium with the cup at \(80^{\circ} \mathrm{C}\). The combination of cup and water is cooled uniformly so that the temperature decreases by \(1.5^{\circ} \mathrm{C}\) per minute. At what rate is energy being removed? Express your answer in watts.

A Styrofoam box has a surface area of \(0.80 \mathrm{~m}^{2}\) and a wall thickness of \(2.0 \mathrm{~cm}\). The temperature of the inner surface is \(5.0^{\circ} \mathrm{C}\), and the outside temperature is \(25^{\circ} \mathrm{C}\). If it takes \(8.0 \mathrm{~h}\) for \(5.0 \mathrm{~kg}\) of ice to melt in the container, determine the thermal conductivity of the Styrofoam.

B. For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for \(24 \mathrm{~h}\) at \(37^{\circ} \mathrm{C}\). A standard constant-temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines, Peace Corps volunteer and MIT engineenAmy Smith invented a low-cost, low-maintenance incubator to fill the need. The device consists of a foaminsulated box containing several packets of a waxy material that melts at \(37.0^{\circ} \mathrm{C}\), interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phasechange material is \(205 \mathrm{~kJ} / \mathrm{kg}\). Model the insulation as a panel with surface area \(0.490 \mathrm{~m}^{2}\), thickness \(9.50 \mathrm{~cm}\), and conductivity \(0.0120 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\). Assume the exterior temperature is \(23.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\) and \(16.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\). (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

8 A \(60.0\) -kg runner expends \(300 \mathrm{~W}\) of power while running a marathon. Assuming \(10.0 \%\) of the energy is deliyered to the muscle tissue and that the excess energy is removed from the body primarily by sweating, determine the volume of bodily fluid (assume it is water) lost per hour. (At \(37,0^{\circ} \mathrm{C}\), the latent heat of vaporization of water is \(\left.2.41 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\right)\)

A. \(100-\mathrm{g}\) cube of ice at \(0^{\circ} \mathrm{C}\) is dropped into \(1.0 \mathrm{~kg}\) of water that was originally at \(80^{\circ} \mathrm{C}\). What is the final temperature of the water after the ice has melted?
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