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Chapter 18: Problem 28

Suppose you mix 7.00 L of water at \(2.00 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) with \(3.00 \mathrm{~L}\) of water at \(32.0^{\circ} \mathrm{C}\); the water is insulated so that no energy can flow into it or out of it. (You can achieve this, approximately, by mixing the two fluids in a foam cooler of the kind used to keep drinks cool for picnics.) The \(10.0 \mathrm{~L}\) of water will come to some final temperature. What is this final temperature?

Short Answer

Expert verified
Answer: The final temperature of the 10.0 L of water after mixing is 23.6°C.

Step by step solution

01

Calculate the mass of each water sample

First, we need to calculate the mass of each water sample. Since the density of water is approximately 1 g/mL (or 1 kg/L), we can directly multiply the volume of each water sample by 1 kg/L to get the mass. Mass of the first water sample (m1): \(m_1 = 7.00 \mathrm{~L} \times 1 \mathrm{~kg/L} = 7.00 \mathrm{~kg}\) Mass of the second water sample (m2): \(m_2 = 3.00 \mathrm{~L} \times 1 \mathrm{~kg/L} = 3.00 \mathrm{~kg}\)
02

Set up the heat transfer equation

Next, we need to set up the heat transfer equation. Since the heat gained by one body (Q1) is equal to the heat lost by the other body (Q2) in an insulated system, we can write the equation as: \(Q_1 = Q_2\) When a substance gains or loses heat, we can calculate the heat transfer (Q) using the formula: \(Q = mc(T_f - T_i)\) Where: - Q is the heat transfer - m is the mass of the substance - c is the specific heat of the substance - T_f is the final temperature - T_i is the initial temperature Since the specific heat of water (c) is the same for both samples, we can write the heat transfer equation as: \(m_1c(T_f - T_{i1}) = m_2c(T_{i2} - T_f)\)
03

Solve for the final temperature (Tf)

Now we have the equation: \(7.00 \times 4.186 \times 10^3 (T_f - 20.0) = 3.00 \times 4.186 \times 10^3 (32.0 - T_f)\) First, we can simplify the equation by dividing both sides by \(4.186 \times 10^3\): \(7.00 (T_f - 20.0) = 3.00 (32.0 - T_f)\) Now, distribute the coefficients: \(7.00T_f - 140.0 = 96.0 - 3.00T_f\) Move all terms related to T_f to one side and constants to the other: \(7.00T_f + 3.00T_f = 96.0 + 140.0\) Combine like terms: \(10.00T_f = 236.0\) Finally, divide both sides by 10.00 to find T_f: \(T_f = 23.6^{\circ} \mathrm{C}\)
04

Conclusion

The final temperature of the 10.0 L of water after mixing the two samples is 23.6°C.

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Most popular questions from this chapter

A \(100 .\) mm by \(100 .\) mm by 5.00 mm block of ice at \(0^{\circ} \mathrm{C}\) is placed on its flat face on a 10.0 -mm-thick metal disk that covers a pot of boiling water at normal atmospheric pressure. The time needed for the entire ice block to melt is measured to be \(0.400 \mathrm{~s} .\) The density of ice is \(920 . \mathrm{kg} / \mathrm{m}^{3} .\) Use the data in Table 18.3 to determine the metal the disk is most likely made of

Water is an excellent coolant as a result of its very high heat capacity. Calculate the amount of heat that is required to change the temperature of \(10.0 \mathrm{~kg}\) of water by \(10.0 \mathrm{~K}\). Now calculate the kinetic energy of a car with \(m=1.00 \cdot 10^{3} \mathrm{~kg}\) moving at a speed of \(27.0 \mathrm{~m} / \mathrm{s}(60.0 \mathrm{mph}) .\) Compare the two quantities.

The thermal conductivity of fiberglass batting, which is 4.0 in thick, is \(8.0 \cdot 10^{-6} \mathrm{BTU} /\left(\mathrm{ft}^{\circ} \mathrm{F} \mathrm{s}\right) .\) What is the \(R\) value (in \(\left.\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\right) ?\)

Several days after the end of a snowstorm, the roof of one house is still completely covered with snow, and another house's roof has no snow cover. Which house is most likely better insulated?

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?
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