Question

To what temperature would you have to heat a brass rod for it to be 1.5 % longer than it is at 28 ∘C?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Change in length due to increase in temperature is given by:

dL = L0*alpha*dT

L0 = original length

dL = 1.5% of L0 = 0.015*L0

dL = 0.015*L0

dT = Change in temperature = Tf - Ti

Ti = Initial temperature = 28 C

Tf = final temperature = ?

alpha = linear expansion coefficient of brass = 19*10^-6

Using these values

dT = dL/(L0*alpha)

dT = 0.015*L0/(L0*19*10^-6) = 0.02*10^6/19

dT = 789.5 C

Tf = Ti + 789.5 = 20 + 789.5

Tf = final temperature = 809.5 C

In two significant figures

Tf = 810 C = 8.1*10^2 C

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