Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
9th Edition
ISBN: 9781305116429
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 20, Problem 20.61AP

An aluminum rod 0.500 m in length and with a cross sectional area of 2.50 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 3(H) K. (a) If one-halt of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool, (b) If the circular surface of the upper end of the rod is maintained at 300 K. what is the approximate boil-off rate of liquid helium in liters per second after the lower half has reached 4.20 K? (Aluminum has thermal conductivity of 3 100 YV/m · K at 4.20 K; ignore its temperature variation. The density of liquid helium is 125 kg/m3.)

(a)

Expert Solution
Check Mark
To determine

The volume of helium boil off by the time the inserted half cools is 4.20K .

Answer to Problem 20.61AP

The volume of helium boil off by the time the inserted half cools to 4.20K is 17.2L .

Explanation of Solution

Given info: The length of the aluminum rod is 0.500m , the cross sectional area of the aluminum rod is 2.50cm2 , the initial temperature of the aluminum rod is 300K and the final temperature of the aluminum rod and helium is 4.20K .

Write the expression for the mass of the substance.

m=ρV (1)

Here,

ρ is the density of the aluminum.

V is the volume of the rod.

The expression for the volume of the rod is,

V=Al

Here,

A is the area of the rod.

l is the length of the rod.

Substitute Al for V in above equation.

m=ρAl

For the mass of the aluminum rod:

Substitute mAl for m , 2700kg/m3 for ρ and 0.500m for l and 2.50cm2 for A in the above equation.

mAl=(2700kg/m3)(2.50cm2×104m21cm2)(0.500m2)=0.16875kg

Thus, the mass of the aluminum rod is 0.16875kg .

For the mass of the helium:

Substitute mHe for m , and 125kg/m3 for ρ in the equation (1).

mHe=125kg/m3VHe

Here,

VHe is the volume of the helium.

Write the expression for the amount of heat lost by the aluminum rod due to the change in temperature.

QAl=mAlcAl(TiTf)

Here,

mAl is the mass of the aluminum rod.

cAl is the specific heat of the aluminum rod.

Tf is the final temperature of the aluminum rod.

Ti is the initial temperature of the aluminum rod.

Substitute 0.16875kg for mAl , 900J/kgK for cAl , 4.20K for Tf and 300K for Ti in above equation.

QAl=(0.16875kg)(900J/kgK)((300K4.20K))=44924.625J

Thus, the heat lost by the aluminum rod is 44924.625J .

Write the expression for the heat gained by the helium.

QHe=mHeLHe

Here,

mHe is the mass of the helium.

LHe is the latent heat of the helium.

Substitute 125kg/m3VHe for mHe and 2.09×104J/kg for LHe in the above equation.

QHe=(125kg/m3VHe)(2.09×104J/kg)=2612500VHe

From the law of the conservation of energy, the heat lost is equal to the heat gained.

Write the expression for the heat lost by the aluminum rod is equal to the heat gained by the helium.

QHe=QAl

Substitute 2612.5VHe for QHe and 44924.625J for QAl in the above equation.

2612500VHe=44924.625JVHe0.0172m3×1000L1m3=17.2L

Thus, the volume of helium boil off by the time the inserted half cools to 4.20K is 17.2L .

Conclusion:

Therefore, the volume of helium boil off by the time the inserted half cools to 4.20K is 17.2L .

(b)

Expert Solution
Check Mark
To determine

The boil off rate of liquid helium when the lower half reached 4.20K .

Answer to Problem 20.61AP

The boil off rate of liquid helium when the lower half reached 4.20K is 0.351L/s .

Explanation of Solution

Given info: The length of the aluminum rod is 0.500m , the cross sectional area of the aluminum rod is 2.50cm2 , the initial temperature of the aluminum rod is 300K and the final temperature of the aluminum rod and helium is 4.20K .

Write the expression for the rate at which energy is supplied to the rod.

P=kA(TiTfdx)

Here,

k is the thermal conductivity of aluminum.

A is the cross sectional area of the aluminum rod.

dx is the change in length of the aluminum rod.

Substitute 3100W/m for k , 2.50cm2 for A , 0.5002m for dx , 4.20K for Tf and 300K for Ti in above equation.

P=(3100W/m)(2.50cm2×104m21cm2)(300K4.20K0.5002m)917W

Thus, the rate at which energy is supplied to the rod is 917W .

Write the expression for the boil off rate of liquid helium after the lower half reaches 4.2K .

R=PρLHe

Substitute 917W for P , 125kg/m3 for ρ and 2.09×104J/kg for LHe in above equation.

R=917W125kg/m3×2.09×104J/kg3.51×104m3/s×1000L1m3=0.351L/s

Conclusion:

Therefore, the boil off rate of liquid helium when the lower half reached 4.20K is 0.351L/s .

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Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)

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