A 1.00 kg ball is at rest on the floor in a 2.10 m ×2.10 m × 2.10 m room of air

Question

A 1.00 kg ball is at rest on the floor in a 2.10 m ×2.10 m × 2.10 m room of air at STP. Air is 80% nitrogen (N2) and 20% oxygen (O2) by volume.

Part A

What is the thermal energy of the air in the room?

Express your answer with the appropriate units.

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Part B

What fraction of the thermal energy would have to be conveyed to the ball for it to be spontaneously launched to a height of 1.50 m ?

Part C

By how much would the air temperature have to decrease to launch the ball?

Express your answer with the appropriate units.

A 1.00 kg ball is at rest on the floor in a 2.10 m ×2.10 m × 2.10 m room of air at STP. Air is 80% nitrogen (N2) and 20% oxygen (O2) by volume.

Part A

What is the thermal energy of the air in the room?

Express your answer with the appropriate units.

E =

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Part B

What fraction of the thermal energy would have to be conveyed to the ball for it to be spontaneously launched to a height of 1.50 m ?

Part C

By how much would the air temperature have to decrease to launch the ball?

Express your answer with the appropriate units.

T =

Explanation / Answer

Given,

m = 1 kg ; dimension = 2.1 x 2.1 x 2.1 meter^3

A)The thermal energy will be:

E = E(N2) + E(O2)

E = 5/2 N(N2) Kb T + 5/2 N(O2) Kb T = 5/2 N kb T

N will be given by

N = p V/ Kb T = 101300 Pa x 2.1 x 2.1 x 2.1 m^3/1.38 x 10^-23 x 273 = 2.49 x 10^26

Eth = 5/2 x 2.49 x 10^26 x 1.38 x 10^-23 x 273 = 2.35 x 10^6 J

Hence, E = 2.35 x 10^6 J

b)We know that, PE = m g h

PE = 1 x 9.8 x 1.5 = 14.7 J

So the fraction would be:

f = 14.7/2.35 x 10^6 = 6.26 x 10^-6

Hence, f = 6.26 x 10^-6

c)for this

dE/dT = E/T

dT = dE/E x dt

dT = 14.7/2.35 x 10^6 = x 273 = 0.00171

Hence, dT = 0.00171 deg C

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