Physics 102 Homework set 5 Solutions L. Weinstein Exercises: ---------- 16.42: You do not need a medium to transfer heat by radiation. Radiation can travel through a vacuum (and it does when it travels from the Sun to the Earth). 16.49: There is a large temperature difference between the hot coffee and the room. Therefore during the time before you drink it, the hot coffee loses heat rapidly by radiation and conduction to the air. If you add the cream quickly, then you lower the temperature of the coffee early. Then, while you are waiting to drink it, there is a smaller temperature difference between the coffee and the air and therefore the coffee will lose much less heat to the air. Therefore, the coffee plus cream will have more total energy (and therefore be at a higher temperature) when you drink it if you add the cream early. 18.2: It does not matter whether the Sun's temperature is 10^7 degrees Kelvin or 10^7 degrees Celsius. There is only a difference of 273 degrees between the two scales: T(Kelvin) = T(Celsius) + 273. 10^7 + 273 = 10^7. 18.4: No, heat does not always flow from an object with larger internal energy to an object with smaller internal energy. A tiny object can have a high temperature and a very small internal energy. A large object can have a small temperature and a large internal energy. Heat will flow from the hotter object to the colder one. 18.6: If you vigorously shake a can of liquid for a few minutes, you will raise its temperature because you are adding energy to the liquid. 18.8: The internal energy of the gas will increase by Delta U = Q - W = 100 J - 80 J = 20 J 18.14: The ultimate source of the energy in wood, coal and oil is solar energy. Oil and coal are nonrenewable because those deposits were formed over millions of years. Trees are considered renewable because they grow in a much shorter time span. 18.16: Transferring 10 calories of energy from the cold lake to the hot cup of tea would not violate the first law of thermodynamics (since energy is still conserved). It would violate the second law, which states that heat does not flow spontaneously from cold to hot. 18.26: No, you cannot cool a kitchen by openning the refrigerator door. The refrigerator does work to pump heat from its inside to its outside. It takes an amount of heat Q from its inside and deposits an amount of heat Q+W (where W was the work it had to do in order to pump the heat) to the outside. If the refrigerator door is open, then you are adding a net amount of heat W to the kitchen and it will warm up. 18.28: You are cooled by a fan on a hot day by convection. The moving air moves hot air away from your body. Think of it as a nice form of windchill. 18.48: a) The chance of 2 coins coming up both heads is (1/2)*(1/2) = 1/4. In ten minutes I can throw a pair of coins dozens of times so two heads will almost certainly appear at least once. b) The chance of ten coins coming up all heads is (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1 / (2^10) = 1/1024 ~ 10^(-3) or one in a thousand. In an hour I can throw 10 coins about 1000 times (since there are 3600 seconds in an hour). Therefore there is a reasonable chance that I will see ten heads appear. c) If I throw 10,000 coins, then the chance that all 10,000 will come up heads is (1/2)*(1/2)*(1/2)*... = 1/(2^10,000) ~ 10^(-3000). This means that all of humanity could throw 10,000 coins repeatedly for the entire age of the universe and the coins will never ever come up all heads. (The universe itself is only 10^17 seconds old.) Problems: ------------------ 18.2: efficiency = (T_hot - T_cold) / T_hot = (2700 K - 270 K) / 2700 K = 0.9 18.6: The plant dumps 1.5*10^8 W of heat. (The efficiency and the power output are not needed for this problem. We are just concerned with the waste heat.) Since Q = c * (delta T) * m energy = specific heat capacity * temperature difference * mass of water we have that power = specific heat capacity * temperature difference * mass of water/second or that mass of water/second = power / (specific heat capacity * temperature difference) Now we know that the specific heat capacity of water is c = 1 cal/(g K) * (4.2 J / 1 cal) * (1000 g / 1 kg) = 4200 J / (kg K) so that mass of water/second = 1.5*10^8 W / ((4.2*10^3 J/(kg K)) * 3 K) = 1.2*10^4 kg/s This is 12 tons per second or 12 m^3 of water per second. (Note that I used heat capacity in J/(kg K). Since 1 degree K is the same size as 1 degree C, this is OK. Estimation: ---------------- The power plant produces 10^9 W of electricity with an efficiency of 1/3. Since the work per second (Work/s) = 10^9 W and epsilon=1/3, this means that it needs a heat input per second of Q per second = (Work/s) / epsilon = 3*10^9 W Now, in one year, it needs a heat input of Q = 3*10^9 W * 1 year * (365 day / 1 yr)*(24 hours / 1 day) * (3600 s / 1 hour) = 10^17 J Burning coal produces 10 J/kg so we need a mass of coal: m = 10^17 J / 10^7 J/kg = 10^10 kg = 10^7 tons This is 10^5 100-ton railroad cars. That is 100 thousand large railroad cars or three 100-car trains a day! If we use uranium instead, then we would need to fission m = 10^17 J / 10^14 J/kg = 10^3 kg = 1 ton. Much less! (In reality we would need about 20 tons of uranium since the isotope that fissions is only a small fraction of the total.)