EART 80 C PROBLEM SET # 1 KEY
Assigned: Monday July 28th
Due: Monday August 4th, 5 pm.
1. (10 pts) Let’s say that the International Space Station has an interior pressure equal to that at sea level on Earth, 100 kPa. In building the ISS, the designers want to calculate how many bolts are needed to hold a small door to outer space in place. Assume that the door is rectangular with dimensions 2 m high x 1.5 m wide.
(a) What is the pressure on the inside of the door (i.e. the pressure wanting to push the door outwards)? What is the pressure on the outside of the door? [2 pts]
What is the pressure on the inside door? This is given above – 100 kPa
On the outside of the door? This is outer space – There is no air, and therefore no molecules to bang against the door from the outside, and therefore the outside pressure is 0.
(b) What, then, is the force on the inside of the door? What is the force on the outside of the door? [3 pts]
Pressure = Force/Area à Force = Pressure * Area
F on inside = Pressure on inside * Area of door = 100x103 Pa * (2 m* 1.5 m) = 3x105 N
F on outside = Pressure on outside * Area of door = 0 Pa * (2 m * 3 m) = 0 N
(c) If each bolt can hold 35 kN of force before breaking, then what is the minimum number of bolts needed to make sure this door holds? [3 pts]
Number of Bolts = (Total force on door) / (Amount of force each bolt holds)
Number of Bolts = (2x105 N) / (3.5x 104 N) = 6.7
Because it would be silly to use only half a bolt, round up to 7 bolts.
(d) Does the answer depend on the orientation of the ISS, i.e. if the door points at Earth, or away from Earth, or in any other direction? Why? [2 pts]
No. The ISS is a sealed container – no matter which way it points, the pressure differences between the inside and the outside of the container are the same. Gravity does affect the station (this is the reason why the station orbits the Earth). But gravity is not responsible for the pressure force – the molecules on the inside of the space station are, which don’t change as a function of the space station orientation.
2. (15 pts) Read the attached story about Larry. We are going to calculate how many balloons Larry really needed in order to “level off at 11,000 feet”. First, let’s say that Larry and all his stuff (beer, lawn chair, gun, sandwiches, balloons) weighs 100 kg.
(a) What must be true for an object to be neutrally buoyant in a fluid (i.e. neither rising nor sinking)? (2 pts)
An object that is less dense than the surrounding fluid will rise. An object that is more dense than the surrounding fluid will sinik. To Neither rise nor sink, then , the object has to have the same density as the surrounding fluid. [In this case, the fluid is air (not water!). By fluid, people mean a medium that flows, so all gases and liquids are fluids.]
(b) Calculate (using the ideal gas law) the density of air at 11,000 ft. (4 pts)
The ideal gas law can be expressed as: PM =ρRT, where P is pressure in units of Pa, M is molecular weight of the gas in kg/mol, ρ is density in units of kg/m3, R is the ideal gas law constant (always equals 8.314 J/mol K in SI units, and T is temperature in units of Kelvin K.
Rearrange the equation to calculate density: ρ = PM/RT
M is that for air, which was given in glass as 29 g/mol = 0.089 kg/mol.
T is 10°C = 283 K
P is 70 kPa, which is 70x 103 Pa or 7.0x104 Pa
So we now have everything we need to calculate density:
ρ = PM/RT = (7.0x 104 Pa) * ( 0.029 kg.mol)/(8.314 J/mol K) * (283 K) = 0.863 kg/m3
(We know that because we are using ONLY SI units for P, M, R and T, then the result for density will be in SI units, which is kg/m3.)
(c) What, then, must the average density of Larry + balloons + other stuff be in order for Larry + balloons + other stuff to float just above the ground? (1 pt)
From (a), we know that the density of a floating object (i.e. neither rising nor sinking) is equal to that of the fluid. Therefore, the overall density of Larry + balloons + other stuff must be 0.863 kg/m3.
(d) Calculate the volume that must be occupied by Larry + balloons + other stuff. (3 pts)
We know that density = mass / Volume, which can be rearranged as Volume = mass/density
We know that the density of Larry + balloons + stuff = 0.863 kg/m3 (from c)
We know that the mass of Larry + balloons + stuff = 100 kg (given)
Therefore, we can calculate the volume of Larry + balloons + stuff as:
Volume of Larry + balloons + stuff = (mass of L + b +s ) / (density of L + b + s) = (100kg)/(0.863 kg/m3) = 116 m3
(e) Assuming that Larry + other stuff have negligible volume, calculate the number of spherical balloons each with a diameter of 4 ft (radius = 0.6 m) that Larry should have used. Recall that the volume of a sphere is given by:
where R is the radius of the sphere. (4 pts)
So we know that the volume of Larry + balloons + stuff is 116 m3, from part (d). We assume that Larry + stuff have negligible volume compared to the balloons. Therefore,
Volume of all balloons = 116 m3.
But the volume of all balloons = (volume of one balloon) * (number of balloons)
But we don’t yet know R. Radius = ½ diameter = ½ * 4 ft = 2 ft = 0.61 m (see conversion below)
Therefore, volume of a single balloon:
0.95 m3
[Note that (0.6)3 is equal to 0.6*0.6*0.6, NOT 3*0.6]
So now we have enough information to calculate the number of balloons. Rearrange the above equation to get:
Number of balloons = (volume of all balloons) / (volume of one balloon) = (116 m3) / (0.95 m3)
Therefore, Number of balloons = 122 balloons.
(f) Given that Larry used 45 balloons, do you think the story is possibly true or an urban myth? (1 pts)
Based on our calculations, Larry needed 2.7 times (=122/45) as many balloons to level off at 11,000 ft, so we would conclude that the story is an urban myth. It may still be true, but the point is that the story as reported doesn’t seem to be very reasonable.
3. (10 pts) Consider a pot of water that contains 7 L of water at 25°C.
(a) Calculate the amount of energy that is needed to bring this water just to a boil. ( 2 pts)
Initially the water is at 25°C. We want to bring the water just to a boil, which is a temperature of 100°C. SO we want to raise the temperature of the water by 75°C. (Note: if you are considering a change in temperature, 75°C = 75 K)
Amount of energy needed = (Heat capacity of water) * (change in T) * (mass of water)
Heat Capacity = 4.2 kJ/(kg K) = 4.2x 103 J/(kg K)
Change in T = 75 K
Volume of water = 7 L * (1 m3 / 1000 L) = 7x 10-3 m3
But the density of water = 1000 kg/m3 (you can look this up).
Therefore, M = ρ * V à mass of water = density * volume = 1000 kg/m3 * 7x10-3 m3 = 7 kg
And therefore,
Energy = 4.2x103 J/(kg K) * 75 K * 7 kg = 2.21x106 J
(b) Find the energy necessary to evaporate all of the liquid. (2 pts)
You need to use the latent heat of evaporation: 2.5x106 J/kg
Energy = (latent heat)* (mass of water) = 2.5x106 J/kg * 7 kg = 1.8x107 J
Note that this is roughly 10 times the value for (a), meaning that the condensation of water vapor releases a lot of energy!
(c) From the time the pot is put on the burner until it is completely dry, find the amount of time required if the burner outputs 2500 W. (3 pts)
First calculate the total amount of energy needed to heat up the water AND evaporate it (i.e., add your answers form part (a) and part (b) à
Total energy = 0.22x107 J + 1.8*107 J = 2.0x107 J (note significant figures here!)
But Power = energy/time
Power = 2500 W (or J/s)
Therefore,
Time = energy/power = 2.0x107 J / 2500 W = 8.0x103 s
This turns out to be: 2 hrs (133 min)
(Note: Your answers may vary somewhat depending on how many digits you used in your calculation).
(d) Calculate the temperature of the burner if it is left on at its full power at 2000 W and it loses all of it’s radiation to it’s surroundings. (3 pts)
(If you gave a halfway decent explanation of your logic on this problem, you received credit)
Use the Stefan-Boltzman Law, which describes how much power is radiated from an object:
Power = A * σ * T4
Power = 2500 W
σ = 5.67x10-8 W/(m2K4)
A = not known
T = what you want to solve for
So we need an area, A. Let’s estimate this value. Let’s say that the burner has a diameter of 8 inches, about half of which is comprised of the burner metal, the rest is air.
A circle 8 inches in diameter has a radius of:
Area of this circle = πR2 = 3.14 * (0.102 m)2 = 3.27x10-2 m2
So the are of the burner = 50% of 3.27x10-2 m2 = 1.63x10-2 m2
Therefore,
Power = 2500 W = 1.63x10-2 m2 * 5.67x10-8 K/(m2*K4) * T4
Solve for T à T = 1.28x103K which corresponds to 1007°C.
[Note that answers will vary depending on the area estimated!]
4. (5 pts) If the Earth’s tilt were 30° rather than 23.5°, discuss whether or not the following would change, how they would change, and why. Draw diagrams for each.
(a) Summertime and wintertime temperatures (1 pt)
If the tilt of the Earth is increased, then the difference between sunlight received by the NH and SH around the solstices is increased. This means that summertime temperatures would be warmer (more sun), while wintertime temperatures would be colder (less sun).
(b) Position of the sun during the equinoxes (1 pt)
At the equinox, the sun would remain in exactly the same place (directly overhead the equator).
(c) Length of day during the solstices (1 pt)
At the summer solstice, the length of the day would increase, while at the winter solstice, the length of the day would decrease (by the same amount).
(d) Position of the arctic circle (2 pt)
The position of the Arctic Circle would move towards the equator, and in fact it would be at 90° – 30° = 60°.
5. In class, we did the calculation for a “perfect” greenhouse gas layer, and this led to an average surface temperature for the Earth of 303 K. The actual average surface temperature for the Earth is about 290 K.
(a) In order to maintain this surface temperature, how much energy (in units W/m2) that is emitted by the greenhouse gas layer is absorbed by the Earth’s surface?
In class, we found that the energy balance for 1 m2 of Earth’s surface can be written as:
Sunlight In + GHG Layer In = Reflected Sunlight out + IR emission out
Where:
Sunlight In = 343 W/m2 * 1 m2 = 343 W
Reflected Sunlight Out = Area * albedo * Sun In = 1 m2 * 0.3 * 343 W = 103 W
IR emission out = AσTsurface4 = 1 m2 * 5.67*10-8 W/m2K4 * (290 K)4 = 401 W
Therefore, actual GHG layer in = 103 W + 401 W – 343 W = 161 W
(b) It is estimated that the greenhouse gases added to our atmosphere due to human activities over the last few centuries have led to an increase of 2.5 W/m2 absorbed by the Earth’s surface from the greenhouse gas layer. Estimate the increase in Earth’s surface temperature during this time period because of this effect.
This time, instead of wanting to solve for “GHG Layer In”, we seek to solve for Earth’s surface temperature, which means finding out what “IR emission out” is. So, as before:
IR emission out = 343 W + (161 W + 2.5 W) – 103 W = 403.5 W
(Note that GHG Layer In is the 161 W we found in (a) plus the extra 2.5 W/m2 caused by GHG due to human activities)
Now, solving for Tsurface:
Tsurface = (P/Aσ)0.25 = (403.5 W/ (1 m2 * 5.67x10-8 W/m2K4))0.25 = 290.4 K
Therefore, the temperature difference is 290.3 K - 290 K = 0.4 K.
We will discuss later is class why this is NOT the observed temperature change during this time period.