A bridge made from concrete (a=12x10 -6 1/ o C ) is exactly 50m long when measur
ID: 1268641 • Letter: A
Question
A bridge made from concrete (a=12x10-6 1/oC ) is exactly 50m long when measured at a temperature of 60 degrees Fahrenheit.
1) What is the length of the bridge on a hot day of 100 degrees Fahrenheit?
L = 49.977 m
L = 49.991 m
L = 50.013 m
L = 50.056 m
2) What is the length of the bridge on a cold day of -10 degrees Fahrenheit?
L = 49.977 m
L = 49.991 m
L = 50.013 m
L = 50.056 m
3) If the bridge were made of aluminum, how would the contraction and expansions change?
greater
less
the same
4) A hole is drilled in the bridge. Is the hole bigger or smaller on the cold day compared to the hot day?
bigger
smaller
the same size
5) A block of lead has a mass of 5kg and a volume of 4.425x10-4m3 at 10 degrees Celsius. What is the density of the block at 200 degrees Celsius? The Coefficient of Volumetric expansion for lead is 87x10-6 1/oC
11003 kg/m3
11098 kg/m3
11116 kg/m3
6) A block of lead with twice the mass goes through the same temperature change. How does the final density compare to what you just calculated?
greater
less
the same
0.1kg of ice initially at a temperature of -10 degrees Celsius is added to a cup with .5kg of water initially at 20 degrees Celsius. The water and ice are isolated thermally from their surroundings. The specific heat of ice is 2000 J/kgoC, the specific heat of water is 4186 J/kgoC and the latent heat of fusion of water is 33.5x104 J/kg
7) What is the final temperature of the water?
T = 1.0oC
T = 1.8oC
T = 2.5oC
T = 3.1oC
8) How much MORE ice (at -10oC) must be added so the final situation is all WATER at 0oC?
m = .018 kg
m = .034 kg
m = .079 kg
m = .127 kg
9) Now, just a little MORE ice is added. What is the final state of the system?
All ice at 0oC.
All ice below 0oC.
All water at 0oC.
All water above 0oC.
Some ice, some water but all at 0oC.
Some ice, some water but below 0oC.
Some ice, some water but all above 0oC.
1.5 moles of Helium begins at a temperature of 50 K, and a volume of .085 m2.
10) If the temperature is increased how will the volume change?
increase
decrease
remain the same
11) If the pressure is decreased, how will the density change?
increase
decrease
remain the same
12) If the temperature is decreased how will the internal energy change?
increase
decrease
remain the same
13) Calculate the final pressure if the temperature is increased to 300 K and the volume decreases to .05 m2
P = 32,780 Pa
P = 66,110 Pa
P = 74,790 Pa
P = 102,500 Pa
P = 116,900 Pa
14) What is the final internal energy of the gas?
1600 J
2800 J
3700 J
4200 J
5600 J
15) What is the rms velocity of the gas molecules?
522 m/s
867 m/s
1367 m/s
1967 m/s
2530 m/s
Explanation / Answer
1)
60 F = 15.5556 C
100 F = 37.7778 C
length of the bridge on a hot day of 100 degrees Fahrenheit = 50 + (12 *50 * (10^-6) * (37.7778 - 15.5556)) = 50.013 m
L = 50.013 m
2)
-10 F = -23.3333 C
length of the bridge on a hot day of 100 degrees Fahrenheit = 50 + (12 *50 * (10^-6) * (-23.3333 - 15.5556)) = 49.977 m
L = 49.977 m
3)
If it was aluminium,
the thermal expansion coeffiecient of aluminium is larger than 12 * 10^-6 m/C
so the contraction and expansions would be greater
greater is the answer
4)
smaller
5)
volume at 200 C = (4.425 * (10^-4)) * ( 1+ (87 * (10^-6) * (200-10))) = 4.49814525 * 10^-4 m^3
so density = 5/ (4.49814525 * (10^-4)) = 11116 kg/m^3 is the answer
6)
same
7)
Let T be the final temperature,
heat gain by ice = (2000 * 0.1 * 10) + (0.1 * 33.5 * (10^4)) + (0.1 * 4186* T)
heat lost by water = (0.5 * 4186 * (20 - T))
so, heat gain by ice = heat lost by water
(2000 * 0.1 * 10) + (0.1 * 33.5 * (10^4)) + (0.1 * 4186* T) = (0.5 * 4186 * (20 - T))
final temperature of the water, T = 2.5 C
8)
For final situation is all WATER at 0C
Let m be the mass of ice , T = 0
then
heat gain by ice = (2000 * m * 10) + (m * 33.5 * (10^4))
heat lost by water = (0.5 * 4186 * (20 ))
so, heat gain by ice = heat lost by water
(2000 * m * 10) + (m * 33.5 * (10^4)) = (0.5 * 4186 * (20 ))
m = 0.1179 Kg
How much MORE ice must be added = 0.1179 - 0. = 0.018 Kg
0.018 Kg is the answer
9)
Some ice, some water but all at 0 C.
10)
increase
11)
decrease
12)
decrease
13)
P1*V1 = n*R*T1
P1 * 0.085 = 1.5 * 8.314 * 50
P1 = 7335.882 Pa
P1 * V1 / T1 = P2*V2/T2
(7335.882 * 0.085) / 50 = (P2 * 0.05)/300
P = 74,790 Pa is the answer
14)
final internal energy = (3/2) * n* R* T = (3/2) * 1.5 * 8.314 * 300 = 5600 J
5600 J is the answer
15)
rms velocity of the gas molecules = sqrt ( 3*R* T /M) = sqrt ( 3 * 8.314 * 300/ 0.004) = 1367 m/s
1367 m/s is the answer
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.