#$&*
course Phy 242
February 12 around 2pm. I am RESUBMITTING this assingment. I submitted it a while back and it never showed up in my portfolio.
110124Estimate how much force it would require to raise the water to a height of 60 cm by squeezing the bottle. An average adult can exert, with one hand, a maximum squeezing force equal to about 1/4 of his or her weight. Naturally you are stronger than average, in the same way that all the children are above average, so be sure to account for this.
****
Approx. 200 Newtons
#$&*
Estimate through what average distance that force is exerted as you tighten your grip, and estimate how much work you require to raise that water. How does your result compare with the energy required to increase the temperature of the air in the bottle?
****
I don’t know how to calculate this…
@& The distance couldn't have been more than 5 cm, the radius of a 2-liter bottle, since you didn't squeeze hard enough to make the opposite sides of the bottle come together.
On the other hand the sides probably got at least a centimeter closer together.*@
#$&*
Suppose the plastic walls of the bottle are .07 cm thick. Suppose the contents of the bottle are originally at a stable 300 Kelvin, and the bottle is immersed in water whose temperature is 330 Kelvin. Assuming that for at least a short time after immersion the inside of the bottle remains at 300 Kelvin, while the outside is raised to 330 Kelvin. What is the temperature gradient across the wall of the bottle?
The temperature gradient is just the rate at which temperature changes with respect to position.
Temperature changes by 30 K, while position changes by .7 mm.
What therefore is the temperature gradient, in K / mm?
****
Temperature gradient: 30K/0.7mm=42.85K/mm
#$&*
What would be the temperature gradient in K / m?
****
42 850K/m
#$&*
What do you estimate to be the surface area of a bottle?
Detail your own estimate. The bottle is approximately cylindrical with diameter 6 cm and altitude 20 cm. That's in the rough neighborhood of 500 cm^2, but get your own result.
****
440 cm^2
#$&*
If the contents of the bottle are observed to gain 10 000 Joules of energy in 50 seconds, then at what rate with respect to clock time is energy flowing across the walls of the bottle?
The requested rate is by definition equal to change in energy / change in clock time = 10 000 Joules / 50 seconds = 200 Joules / second.
At what rate is energy flowing, per second, per square centimeter?
****
200 J/s to per sec, per cm^2
(200 J/s)/second = 200 J
(200 J/s)/cm^2 =
I don’t know what I’m doing wrong. I’m getting hung up on this problem and the next one and I’m sure it’s not hard…..
#$&*
@& You've got 200 J / s over the entire surface, which is a lot more than a cm^2.
How many cm^2 did you previously estimate?
How many J / s do you therefore have per cm^2?*@
At what rate is energy flowing, per second, per square centimeter, per unit of temperature gradient?
****
@& Assuming you got the last one, the rate at which energy is flowing, per second, per square centimeter
What is the temperature gradient, and what are its units?
So what is the rate at which energy is flowing, per second, per square centimeter, per unit of temperature gradient?*@
#$&*
That last rate is called the thermal conductivity of the plastic. What in the blazes are the units of that quantity? What are its SI units?
****
Thermal conductivity or flux = (quantity of heat/c.s. area*time) * (thickness/temperature difference)
SI Units: W/(m*K)
#$&*
Look up the thermal conductivity of plastic. Compare with your result (which, even if you do the problem right, isn't likely to be very close, because that 10 000 J in 50 sec is probably not accurate at all).
****
I will have to do this problem when I figure out the 2 problems before the last.
#$&*
How could we use such a system to measure the conductivity of the plastic that makes up the bottle?
****
I’ve heard of a “transient” method, that simply perform a measurement during the process of heating up. (needle probes)
#$&*
If after raising the temperature to 318 K to fill the tube, we raise the temperature another 20 K, how much water would we be able to collect at the 60 cm height?
****
338K total
If V1=100mL, T1=318K, and T2=338K, I solved for V2.
V2=106.3 mL
So, 106.3-100=6.3 mL
#$&*
By how much would the gravitational PE of the water change?
****
‘dPE= (weight) * (center of mass at column height)
=(mass*980cm/s^2) * (30cm)
@& Good, but you know the volume of the water (see preceding), so you can fill in its mass and calculate the PE change.*@
#$&*
How much additional translational KE would be required?
****
3/2nRT
‘dKE_trans = (3/2)*(n)*(8.31 J/mol K)*(318 K)
I know to solve for “n”, you would do n=(PV/RT), but when I use these, P=106 000 Pa, V=100mL, R=8.31J/mol K, and T=318K, I am not getting a reasonable answer. What am I doing wrong?!
#$&*
@& You should get n = .004 moles, roughly, if you're using a 100 ml bottle.
However I suspect that the air in the bottle has more than 100 ml volume.*@
@& The expression you give is 3/2 n R T, which is equal to KE_trans.
It's not equal to `dKE_trans, which you could calculate by finding KE_trans at the two different temperatures and calculating the difference.*@
What is the PE change as a percent of the additional translational KE?
****
10%
#$&*
Most of the air molecules in the container also have rotational KE, which turns out to be 2/3 of their translational KE, on the average. So how much thermal energy must be added to the system between the 300 K and 318 K temperatures, when the volume of the system didn't change by much?
****
#$&*
When the gas is allowed to expand, it must do additional work against atmospheric pressure, equal to 2/3 of the additional translational KE. What therefore is the total thermal energy required to raise the temperature of the system from 318 K to 338 K?
****
#$&*
How much PE did we gain, as a percent of the thermal energy required to raise the temperature of our gas from 300 K to 318 K?
****
#$&*"
@& I'm not sure what happened to my original posting. In any case I've inserted a copy of my original comments (which were archived).*@