# 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.) 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...

## See This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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`