Heat and increasing a rod's length

AI Thread Summary
To determine the energy needed to increase the length of a 1.0m metal rod by 7.5 × 10−3 m, the focus should be on the change in length rather than volume. The rod's mass can be calculated using its density and fixed dimensions, which do not change with heat application. The coefficient of linear expansion is crucial for calculating the necessary temperature increase to achieve the desired length change. While volume will change as length increases, the problem specifically requires addressing the linear expansion. The correct approach simplifies the calculations by concentrating on ΔL and the associated temperature change.
Lisa Marie
Messages
9
Reaction score
0

Homework Statement


A 1.0m long rod of metal has a diameter of 0.75 cm. This metal has a coefficient of linear expansion α = 4.8 × 10−5 1/K , a density of 9.7 × 103 kg/ m3 , and a heat capacity of 390 J/K . How much energy needs to be added as heat to increase the length of the rod by 7.5 × 10−3 m?

Homework Equations


V0=πr2h
ΔV=βV0ΔT
m=vρ

The Attempt at a Solution


V0=πr2]h
=π(0.0075/2)2(1)
=4.42×10-5

V0=πr2h
=π(7.5×10-3+1)(0.0075/2)2
=4.45×10-5

ΔV=βV0ΔT
ΔT=ΔV/(3αV0)
=(4.45×10-5-4.42×10-5)/(3(4.42×10-5))

m=vρ where do I go from here? Because the the mass would always be changing
 
Physics news on Phys.org
The mass is fixed. Probably the density will change.
 
NascentOxygen said:
The mass is fixed. Probably the density will change.

So you would do M=∫ρdV to find the mass?
 
You know the rod's dimensions and its density, so can determine mass. This mass doesn't change when you apply heat to expand it.

I'm wondering why you involve ΔV when all that is required is ΔL? I could be wrong, but it seems that you are making the problem more difficult than is intended.
 
NascentOxygen said:
You know the rod's dimensions and its density, so can determine mass. This doesn't change when you apply heat to expand it.

I'm wondering why you involve ΔV when all that is required is ΔL? I could be wrong, but it seems that you are making the problem more difficult than is intended.

Ok thanks! I'll try it with just change in length but doesn't the volume change since the length is changing?
 
Lisa Marie said:
Ok thanks! I'll try it with just change in length but doesn't the volume change since the length is changing?
Certainly volume will change when length and diameter change. But the problem specs centre on the change in length.
 
NascentOxygen said:
Certainly volume will change when length and diameter change. But the problem specs centre on the change in length.
So I should use this volume:
V0=πr2h
=π(0.0075/2)2(1)
=4.42×10-5
...because I'm not getting the right answer
 
You are attacking the problem backwards. How much of a temperature increase do you need to increase the length by the desired amount? What is the mass of the rod?

Chet
 
Back
Top