Solving for Temp Change of Epoxy Frames for Lenses Insertion

AI Thread Summary
To insert lenses with a radius of 2.21 cm into epoxy frames with a radius of 2.20 cm at room temperature (20.0°C), the frames must be heated to a specific temperature calculated using the coefficient of linear expansion for epoxy, which is 1.30 x 10^-4 (°C)^-1. The discussion clarifies that the unit (°C)^-1 indicates the inverse relationship with temperature, as it appears in the denominator of the thermal expansion equation. The equation used is α = (∆L/L) / T, where the temperature T is in the denominator, resulting in the coefficient having units of (°C)^-1 or K^-1 if Kelvin is used. Understanding this unit is essential for accurately applying thermal expansion principles in practical scenarios. The conversation emphasizes the importance of unit consistency in calculations involving thermal expansion.
KristinaMr
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Homework Statement


This is the problem.

A pair of eyeglass frames is made of epoxy plastic. At
room temperature (20.0°C), the frames have circular
lens holes 2.20 cm in radius. To what temperature must
the frames be heated if lenses 2.21 cm in radius are to
be inserted in them? The average coefficient of linear
expansion for epoxy is 1.30 x 10^-4 (°C)^-1.

Homework Equations

The Attempt at a Solution


I know how to solve this but I'm not very sure what the ℃ elevated to -1 means. Can someone explain what that means and why it is expressed this way?
 
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KristinaMr said:

Homework Statement


This is the problem.

A pair of eyeglass frames is made of epoxy plastic. At
room temperature (20.0°C), the frames have circular
lens holes 2.20 cm in radius. To what temperature must
the frames be heated if lenses 2.21 cm in radius are to
be inserted in them? The average coefficient of linear
expansion for epoxy is 1.30 x 10^-4 (°C)^-1.

Homework Equations

The Attempt at a Solution


I know how to solve this but I'm not very sure what the ℃ elevated to -1 means. Can someone explain what that means and why it is expressed this way?
Look at wiki page, "Thermal expansion coefficients for various materials" section. It's just the inverse of the temperature unit.

Write down the equation of the thermal expansion, and weigh down the units for the thermal coefficient.
 
α=(∆L/L)/T

So the units are ℃^-1 because the temperature in the equation is in the denominator (which means T ^-1) . Right?
 
KristinaMr said:
α=(∆L/L)/T

So the units are ℃^-1 because the temperature in the equation is in the denominator (which means T ^-1) . Right?

Yes. The physical unit for the length is irrelevant because you have (( ΔL/L )) which always cancels out.

The choice of the unit for the temperature influence the unit of your coefficient and vice versa. You can choose to use K (Kelvin) instead of °C (Celsius). In this scenario, α have a unit of K^-1
 
Thank you for your help :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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