HOMEWORK 3 ANSWER SHEET
1. a) Conduction -- the transfer of heat from molecule to molecule within a substance
Convection -- the transfer of heat by the rising of warm fluid and sinking of cold fluid
(Note: Advection is another form of transferring heat. Advection is the horizontal transfer of
some property (in this case, temperature -- internal energy) of air from one area to
another.)
Radiation -- the transfer of heat by electromagnetic waves
b) i) Radiation is the major method of heat transfer warming Bob's face. The hot fire gives off a large amount of radiation, while air is not a good conductor of heat, and Bob does not feel a warm wind indicating that advection is bringing warm air from next to the fire to warm his face.
ii) Conduction is the major method of heat transfer making Bob's hand cold. When two objects of different temperatures, such as the can of Bud and Bob's hand, come into contact, conduction transfers heat from the warmer object to the cooler object. The molecules in Bob's hand are not moving around to transport heat towards the can of Bud, so convetion is not taking place, and Bob's hand is not radiating a great deal of heat into the beer can.
iii) Convection is the major method of heat transfer causing the air at the top of the chimney to be warm. Most of this air was heated by the fire, became less dense, and rose up through the chimney to transport heat there, while the air in the chimney is not a good conductor of heat, and the amount of radiation from the fire reaching the top of the chimney is small compared to the amount of heat being carried there by the rising air above the fire.
2. In each of these problems, energy is being used for some process. In a, energy is being used to evaporate water, while in b and c, energy is being used to increase the temperature of a given mass of air and water respectively.
a) We know (from the problem and from the equation on the back of the homework sheet):
Mass of water evaporated = 4 kg
Latent Heat of Vaporization = 2.5 x 106 J / kg
Energy used for evaporation = (Latent Heat of Evaporation) x (Mass Evaporated)
So, E = 2.5 x 106 J / kg x 4 kg = 107 J
b) We know (from the problem and from the equation on the back of the homework sheet):
Mass of air = 1 x 104 kg
Change in temperature = 1deg. C = 1 K
Atmospheric Specific Heat Capacity = 103 J / (kg K)
Energy absorbed = (Specific heat) x Mass x (Change in Temperature)
So, E = 103 J / (kg K) x 1 K x 1 x 104 kg = 107 J
c) We know (from the problem and from the equation on the back of the homework sheet):
Mass of water = 2.4 x 103 kg
Change in temperature = 1deg. C = 1 K
Specific Heat Capacity of liquid water = 4.2 x 103 J / (kg K)
Energy absorbed = (Specific heat) x Mass x (Change in Temperature)
So, E = 4.2 x 103 J / (kg K) x 2.4 x 103 kg x 1 K = 107 J
d) It takes the same amount of energy to evaporate 4 millimeters of water per unit area, warm 2.5 meters of water per unit area one degree Celsius, and warm a 1 m2 column extending through the entire depth of the atmosphere, one degree Celsius. So, if the ocean is around 4 km deep, it can store MUCH more energy than the atmosphere before observing a 1deg. C change in temperature. Likewise, evaporating a little bit of water into the atmosphere transfers a lot of energy to the atmosphere. If this water vapor is transported somewhere (say, from the tropics to the poles), and then condenses, it will 'release' this energy and warm the air.
3. Batman is not worried because he knows (from his ATMS 101 class) that the atmosphere already absorbs 100% of the outgoing radiation at 19 micrometers. So, if the Joker adds a gas that absorbs in this range, it will not contribute to global warming. However, if there was a typo, then Batman should get off his couch because a gas that absorbs at 9 micrometers will absorb in the atmospheric window. Though there is a spike near 9 micrometers, it does not absorb at 100% efficiency. Also, adding more gas that absorbs near that spike will broaden that spike, so it will have a larger effect. We graded this problem on understanding of the idea, so if you mentioned that there was already a spike there, we didn't count off too much (if any).
4. a) Venus is farther away from the sun, so the energy that the sun emits is spread over a larger area than Mercury. This means that the amount of radiation per unit area recieved at the surface of Venus will be less than that recieved at the surface of Mercury.
If we wanted to go farther and understand why page 41 of the book says that if the distance from the sun is doubled, the amount of energy received per unit area is quartered, then we can refer to the geometry above. Notice that the sun is emitting the same amount of energy in all directions. So, if we want to see how much energy per unit area is recieved at a distance R, then we need to look at how much area this energy is spread out over at that distance. Noting that the surface area of a sphere (from the back of the homework) is: Area = 4 x [[pi]] x R2 , where [[pi]] = 3.14, we can write the energy per unit area as:
Energy per unit area = (Total Energy) / (Total area)
= E / (4 x [[pi]] x R2)
So, the amount of energy recieved per unit area decreases with the sqare of the distance.
5. a) 30% of 342 W / m2 is reflected, so 70% must be absorbed.
So, 342 W / m2 x 0.7 = 239 W / m2 absorbed.
b) Knowing that the amount of radiation absorbed must equal the amount of radiation emitted, the earth must be emitting 239 W / m2.
c) If E = 5.67 x 10-8 x T4, then we can find T by inverting the equation:
T4 = E / (5.67 x 10-8)
So,
T = E_____ = 255 K = -18deg. C
(5.67 x 10-8)
d) If the earth were covered with snow, then the albedo would increase, as snow is a good reflector of incoming solar radiation. This means that less energy would be absorbed by the atmosphere and earth, so the radiative equilibrium temperature should go down.
6. a) Recall from problem 2, and from the back of the homework sheet:
Mass of air = 1 x 104 kg
Change in temperature = 1deg. C = 1 K
Atmospheric Specific Heat Capacity = 103 J / (kg K)
Energy absorbed = (Specific heat) x Mass x (Change in Temperature)
So, (Change in Temperature) = (Energy Absorbed) / (Mass x (Specific Heat))
(Change in Temperature) = (1.9 x 109 J / m2) / (104 kg / m2 x 103) = 190deg. C
b) This is a huge increase in temperature over the course of a year, and has not be observed. Noting that the poles would experience a similar decrease in temperature, there must be some transfer of energy from the tropics to the poles. This happens in the atmosphere and the ocean. From problem 2, we found that if we evaporate water into the air, then the air is storing energy in the form of latent heat. If we then tranfer this water vapor towards the poles, when it condenses again it will release that heat in the form of energy. Likewise, if the sun heats the atmosphere and ocean in the tropics, as that air or water moves to the poles it brings energy with it. This energy is then reradiated to space. Another way to think about what is going on is to imagine what would happen if there were no atmospheric or oceanic circulations. Then, after the earth equilibrated (no temperature increase or decrease), the amount of radiation absorbed by the earth would equal the amount emitted by the earth, and the two curves would be identical. Notice that the tropics would be VERY warm, and the poles VERY cold in this scenario. So, atmospheric and oceanic circulations are acting to warm the poles, and cool the tropics.