Which Will Have The Higher Temperature?
Consider two identical homogeneous spheres made of the same material. Each of the spheres is of mass
m
and radius
r
. Both have the same initial temperature. One of them rests on a thermally insulating horizontal plane and the other hangs from an insulating thread. Now equal amount of heat
Q
is given to each of the spheres as a result of which their temperature rises.
Find the absolute value of the difference in the final temperatures of the spheres in Kelvins rounded off to the nearest integer.
Details and Assumptions:
- All kinds of heat losses are negligible.
- m = 1 kg
- r = 1 m
- g = 1 0 m/s 2
- Q = 1 0 0 J
- Coefficient of linear expansion of the material ( α )= 1 0 − 6 K − 1
- Specific heat capacity of the material of the spheres ( C ) = 2 × 1 0 − 2 J K − 1 kg − 1
The answer is 5.
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4 solutions
Oh dammit, I took the radius of 1m to be a diameter of 1m and got to the solution 2.5K :(
Why you didn't calculate the difference in linear dimensions based on initial temperature? In your solution there is implicitly acknowledged that initial temperature is equal to 0 Celsius.
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Well, △ h = r α △ T for any initial T , I never assumed it to be 0 ∘ C
△ T = ( m c ) 2 − ( m g α r ) 2 2 m g α r Q
this is the IPhO 1967 problem
i got exactly 5 , am i wrong ? since the problem author sounds like the answer will be a fraction and it was to be rounded off . :{|}
U shouldnt have got
Let Δ T be the increase in temperature of a sphere.
The amount of work done in expansion (lowering or elevating the sphere) is W = m g Δ h = ± m g α Δ T R = ± 1 0 − 5 Δ T . The increase in thermal energy is Δ U = m c Δ T = 2 × 1 0 − 2 Δ T .
Together, these changes account for the heat. Therefore Q = Δ U − W = ( 2 × 1 0 − 2 ± 1 0 − 5 ) Δ T = 1 0 0 , so that Δ T = 2 × 1 0 − 2 ± 1 0 − 5 1 0 0 ≈ 5 0 0 0 ∓ 2 . 5 . Thus the temperature difference between the spheres becomes ( 5 0 0 0 + 2 . 5 ) − ( 5 0 0 0 − 2 . 5 ) = 5 K .
On the hanging ball wouldn't the gravity and tension put up equal and opposite work such that entire energy is used up in raising internal energy ?
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In the expansion of the spheres, the forces of gravity and tension/normal force are not quite balanced during for a brief amount of time. That is precisely when their CoM change position and when opposite amounts of work are done on them.
More specifically, for the hanging ball W g − W t = 1 0 − 5 Δ T and for the resting ball W n − W g = 1 0 − 5 Δ T .
Center of mass of the sphere which is kept on plane rises up by △ h = r α △ T ≈ r α C Q while the hanged one gets down by same amount.
Hence, also considering the changes in GPE, the difference in heats utilised in increasing their temperature is:
2 m g △ h
Hence, the difference in temperatures is C 2 m g △ h
= C 2 2 m g r α Q = 5