Student Question

a) A circular copper ring at 20.0°C has a hole with an area of 9.980 cm2.

What minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.000 cm2?

(b) Suppose the ring and the rod are heated simultaneously. What minimum change in temperature of both will allow the ring to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.)

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Initial length of copper ring is

`L_0 =sqrt(S_0/pi) =sqrt((9.98*10^-4)/pi) =0.0178234 m=1.78234 cm` .

Final Length of copper ring is

`L =sqrt(S/pi) =sqrt(10^-3/pi) =0.0178412=1.78412 cm`

Linear coefficient of thermal expansion for copper is

`alpha =16.6*10^-6 K^(-1)`

The law of linear thermal expansion writes as

`L =L_0*(1+alpha*Delta(T))`

`Delta(T) = (1/alpha)*[(L/L_0)-1]=1/(16.6*10^-6)*(1.78412/1.78234-1)=60.16 deg` `T=T_0 +60.16 =20 +60.16 =80.16 Celsius`

b) coefficient of thermal expansion for Steel is

`alpha_(Fe) =11*10^-6 K^-1`

Now the relation between the lengths is

`L_0*(1+alpha_(Cu)*Delta(T)) =L*(1+alpha_(Fe)*Delta(T))`

`L-L_0 = Delta(T)*(L_0*alpha_(Cu) -L*alpha_(Fe))`

`Delta(T) = (L-L_0)/(L_0*alpha_(Cu) -L*alpha_(Fe)) =178.69 Celsius`

`T = T_0 +178.69 =198.69 Celsius`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial