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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 1.19-kg aluminum pot contains 2.31 L of water. Both pot and water are initially at \(19.7^{\circ} \mathrm{C} .\) How much heat must flow into the pot and the water to bring their temperature up to \(95.0^{\circ} \mathrm{C}\) ? Assume that the effect of water evaporation during the heating process can be neglected and that the temperature remains uniform throughout the pot and the water.

Suppose \(0.010 \mathrm{~kg}\) of steam (at \(100.00^{\circ} \mathrm{C}\) ) is added to \(0.10 \mathrm{~kg}\) of water (initially at \(\left.19.0^{\circ} \mathrm{C}\right)\). The water is inside an aluminum cup of mass \(35 \mathrm{~g}\). The cup is inside a perfectly insulated calorimetry container that prevents heat flow with the outside environment. Find the final temperature of the water after equilibrium is reached.

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).
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