Physics 102

Chapter 18 Homework Solutions

L. Weinstein

Exercises:

2: 10^7 degrees C is 10,000,000 degrees C = 10,000,273 degrees K. These two numbers are effectively identical. The 273 degree difference is much less than our uncertainty in the temperature since it is ‘about 10^7 degrees’.

4: Yes, the temperature will increase. By shaking the can, you are imparting kinetic energy to it. This kinetic energy is then converted to thermal energy and raises the temperature of the can. You have to shake A LOT to get the temperature to go up a little.

12: The ultimate source of energy for a hydroelectric power plant is the sun. Heat from the sun evaporates water from lakes and oceans. Some of this water then falls as rain in the mountains. This water flows through streams and rivers to the hydroelectric dam.

14: Transferring heat from the colder lake to the hotter tea does NOT violate the first law of thermodynamics since the total amount of energy does not change. It does violate the second law of thermodynamics since that states that heat will not flow spontaneously from a colder object to a hotter object.

20: The efficiency is (T_hot – T_cold) / T_hot. If T_cold is lowered, then the difference between T_hot and T_cold is increased and the efficiency increases.

26: There are two reasons: 1) the refrigerator must do work to pump heat from the colder object (the inside of the refrigerator) to the hotter object (the kitchen). The bigger the temperature difference, the more work is needed to pump the heat. 2) More heat leaks into the refrigerator when the room is hotter. According to Newton’s Principle, the bigger the temperature difference between hot and cold objects, the more heat flows from hot to cold.

28: It is not wasteful to leave the lights on in a building that is electrically heated. Electrical energy is converted is converted to light and heat in a light bulb. The light from the lightbulb is eventually absorbed (after some number of reflections) and also converted to heat. Thus, almost all of the electrical energy used by the light bulb ends up as heat. Therefore, it does the same thing as the electrical heater which converts all of the electrical energy into heat.

It is very wasteful to leave the lights on in an air conditioned building. You are paying for the electrical energy to run the lights and then you are paying again for the energy to run the air conditioner to remove the heat generated by the light bulbs.

36: The principle of entropy says that disorder increases in isolated systems. The chicken is not an isolated system. It receives food from the outside world.

An analogy is that my office tends to become disordered with time. It only becomes more ordered if I put in energy (in the form of work) and actually clean my office. You can reduce entropy in a system if you do work on it from the outside.

38: For this problem you need to calculate probabilities. The chance of 1 coin coming up heads is ½. The chance of 2 coins coming up heads is ½*½ = ¼ = (½)^2. The chance of 3 coins coming up heads is (1/2)^3. The chance of N coins coming up heads is (1/2)^N.

a) Since the chance of 2 coins coming up heads is ¼, I will have to throw them 4 times on average before they come up all heads. I can easily do this in 10 minutes.

b) The chance of 10 coins coming up heads is (1/2)^10 = 1/1024. I can probably throw ten coins 1000 times in an hour (which has 3600 seconds).

c) c) The chance of 10,000 coins coming up heads is (1/2)^10,000. We know that (1/2)^10,000 = [ (1/2)^10 ]^1000 = (1/1024)^1000 = (10^(-3))^1000 = 10^(-3000). This is an amazingly tiny number. For comparison, 1 trillion = 10^12. Also, the entire universe is only about 10^18 seconds old. This means that, even if someone else was handing you the coins, examining them to see if they are all heads, and picking them up so that all YOU have to do is throw the coins, there is no way you could do it. In practice, 10,000 coins will NEVER come up all heads during the entire life of the universe.

Problems:

2: The ideal efficiency is epsilon = (T_hot – T_cold) / T_hot = (2700 – 270)/2700 = 0.9

6: The electrical power and the efficiency are irrelevant. All we need to know is that the power plant needs to get rid of 1.5*10^8 W of heat. (Remember that 1 W = 1 Joule/second.) It transfers the heat to water. Each kg of water can take 4184 J/C * 3C = 1.2*10^4 J. Each second the power plant transfers 1.5*10^8 J. This means that each second it needs 1.5*10^8 J / 1.2*10^4 J/kg = 1.2*10^4 kg.

More formally, you need

Flow = 1.5*10^8 J/s / (4.2*10^3 J/kg-C * 3 C) = 1.2*10^4 kg/s

Estimation:

To generate 1 GW of power with an efficiency of 1/3, we need an input power of 1 GW / 1/3 = 3 GW = 3*10^9 W. Now we need the number of Joules used in 1 year so we need to find the number of seconds in 1 year. 1 year = 365 days/yr * 24 hours/day * 60 minutes/hr * 60 seconds/minute = 3.1*10^7 s. This means that the power plant uses 3*10^9 W * 3.1 * 10^7 s = 10^17 J in one year.

Now, to get 10^17 J, you need 10^17 J / 10^7 J/kg = 10^10 kg of coal.

10^10 kg = 10^7 tons = 10^5 railroad cars = 300 RR cars per day!

That’s a LOT of coal.

Let’s consider nuclear energy. Now we need 10^17 J / 10^14 J/kg = 10^3 kg = 1 ton. We only need 1 ton of Uranium.