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starmoonsun
19 Jul 07, 16:27

1) a unifrom capillary tube, closed at one end, contained air trapped by a thread of mercury 85mm long. when the tube was held horizontally, the length of the air column was 50mm. when it was held vertically with the closed end downwards, the length was 45mm. find the atm in Nm-2
(take g=10, density of mercury = 14000kg/m^3)

ans= 107000Pa

2) A U-tube with 30cm of water is set up vertically and 12cm of olive oil is poured carefully into the left hand limb of the tube. it is observed that water will be pushed up in the other limb. the density of water and olive oil are 1000kg/m^3 and 920kg/m^3 respectively.
A further 6cm of olive oil is added into the left hand limb. how much further will the water level rise in the right limb?

ans = 2.76mm

3) An airline pilot ascending from take off notes that temp at a height of 2500m falls to 273K. the temp n atm at sea are 287K and 1o1kPa respectively.
a)calculate the atm at height of 2500m.
ans= 98300 Pa
b) the atm at cruising height of 10000m is 26kPa, but the pressure within the airliner maintained at 95kPa. Each window by the passenger seat has an area of 0.12m^2, and is secured by 20 bolts. calculate the tension in each bolt.
ans = 414 N

4) A hot water tank of heat capacity 6000J/K contains 120kg of water at 15 degrees celsius. what is the time taken to raise the temp of water to 50 degrees celsius using a heater coil of power of 2500W, given that the specific heat capacity of water is 4200 K/kg/K?
ans = 7140s.

can anyone help me with the steps and provide explanation? thanks!
hisoka
19 Jul 07, 16:29

Originally posted by starmoonsun:
1) a unifrom capillary tube, closed at one end, contained air trapped by a thread of mercury 85mm long. when the tube was held horizontally, the length of the air column was 50mm. when it was held vertically with the closed end downwards, the length was 45mm. find the atm in Nm-2
(take g=10, density of mercury = 14000kg/m^3)

ans= 107000Pa

2) A U-tube with 30cm of water is set up vertically and 12cm of olive oil is poured carefully into the left hand limb of the tube. it is observed that water will be pushed up in the other limb. the density of water and olive oil are 1000kg/m^3 and 920kg/m^3 respectively.
A further 6cm of olive oil is added into the left hand limb. how much further will the water level rise in the right limb?

ans = 2.76mm

3) An airline pilot ascending from take off notes that temp at a height of 2500m falls to 273K. the temp n atm at sea are 287K and 1o1kPa respectively.
a)calculate the atm at height of 2500m.
ans= 98300 Pa
b) the atm at cruising height of 10000m is 26kPa, but the pressure within the airliner maintained at 95kPa. Each window by the passenger seat has an area of 0.12m^2, and is secured by 20 bolts. calculate the tension in each bolt.
ans = 414 N

4) A hot water tank of heat capacity 6000J/K contains 120kg of water at 15 degrees celsius. what is the time taken to raise the temp of water to 50 degrees celsius using a heater coil of power of 2500W, given that the specific heat capacity of water is 4200 K/kg/K?
ans = 7140s.

can anyone help me with the steps and provide explanation? thanks!
1. what is the temperature of the air that is trapped? it will affect the answer.
eagle
19 Jul 07, 16:41

Originally posted by starmoonsun:
1) a unifrom capillary tube, closed at one end, contained air trapped by a thread of mercury 85mm long. when the tube was held horizontally, the length of the air column was 50mm. when it was held vertically with the closed end downwards, the length was 45mm. find the atm in Nm-2
(take g=10, density of mercury = 14000kg/m^3)

ans= 107000Pa
Let atm = P1
Volume of air trapped is proportional to length of air column

When it is horizontal, only atm is involved
when it is vertical, pressure of air = atm + pressure due to mercury column
= P1 + 14000 * 10 * 0.085 (pgh)
= P1 + 11900

Hence, using P1V1 = P2V2 (Boyle's Law)
P1 (50) = (P1 + 11900) (45)
Solving the equation,

P1 = 107000 Pa (3 sf)
eagle
19 Jul 07, 16:54

Originally posted by starmoonsun:
2) A U-tube with 30cm of water is set up vertically and 12cm of olive oil is poured carefully into the left hand limb of the tube. it is observed that water will be pushed up in the other limb. the density of water and olive oil are 1000kg/m^3 and 920kg/m^3 respectively.
A further 6cm of olive oil is added into the left hand limb. how much further will the water level rise in the right limb?

ans = 2.76mm
Check if the answer is 2.76 cm, not 2.76 mm. I'm rather sure it is 2.76 cm.

Soln:

U can draw a U tube, left side with oil and water, right side with water

let the height of water at the left side = x cm
thus the height of water at the right side = 30 - x cm

Also, height of oil at the left side = 12 cm

Using pgh,

pressure at the bottom due on left side = 12 (920) + x (1000)
pressure at the bottom due on right side = (30 - x) (1000)

Equating both sides of the pressure, x = 9.48 cm

Repeat the above for heigh of oil = 12 + 6 = 18cm
you will get a new x of 6.72 cm

Hence, the difference = 2.76 cm => the water level rose by 2.76 cm on the other side.

--------------------------------------------------------------------------------------------

Alternatively, you can use this method if you understand the concept

when 6 cm of oil is added onto one side, the extra pressure will have to be balanced out; i.e. the pressure on the left side (oil side) will have to decrease by 3cm of oil, whereas the pressure on the right side will have to increase by 3cm of oil.

height of water level that is equal to pressure due to 3cm of oil
= (3 * 10 * 920) / (10 * 1000)
= 2.76 cm
angelfairy
19 Jul 07, 17:10

im guessing eagle will do 3 so i shall do 4:

Question:

4) A hot water tank of heat capacity 6000J/K contains 120kg of water at 15 degrees celsius. what is the time taken to raise the temp of water to 50 degrees celsius using a heater coil of power of 2500W, given that the specific heat capacity of water is 4200 K/kg/K?
ans = 7140s.

Answer:

Q = mct where t is the rise in temperature, so

amount of energy needed for the rise in temperature of the water only, Q = (120)(4200)(50 - 15) = 17640000 J

amount of energy needed for the rise in temperature of the tank only, Q = (6000)(50 - 15) = 210000J

total energy required to be supplied by heater coil = 17640000 + 210000 = 17850000J

since power = energy/time, time needed = energy/power = 17850000/2500 = 7140s (ANS)
eagle
19 Jul 07, 17:26

Originally posted by hisoka:
1. what is the temperature of the air that is trapped? it will affect the answer.
the temperature should be a constant because it is still at the same location; it did not change.

hence it will not affect the answer because according to the gas equation (A level stuff), the temperature will cancel out on both sides
eagle
19 Jul 07, 17:31

Originally posted by starmoonsun:
3) An airline pilot ascending from take off notes that temp at a height of 2500m falls to 273K. the temp n atm at sea are 287K and 1o1kPa respectively.
a)calculate the atm at height of 2500m.
ans= 98300 Pa
b) the atm at cruising height of 10000m is 26kPa, but the pressure within the airliner maintained at 95kPa. Each window by the passenger seat has an area of 0.12m^2, and is secured by 20 bolts. calculate the tension in each bolt.
ans = 414 N
a) using P1/T1 = P2/T2 should give you the answer...

But I got 96kPa here... can someone else confirm?

b)
Note that I changed it to force, not pressure:

Force on window from outside = 26k * 0.12 = 3.12kN
Force on window from inside = 95k * 0.12 = 11.4kN
Difference in force = 11.4 - 3.12 = 8.28kN

This force difference is supplied by the 20 bolts, hence
tension in each bolt = 8280N / 20 = 414 N