- #1
Saladsamurai
- 3,020
- 7
A certain steel rod has a diameter of 3.00 cm at 25 celsius and brass ring has an inner diameter of 2.992 cm at 25 degrees celsius. At what common temperature will the ring just fit over the rod?
I am using rate of linear expansion: [tex]\Delta d=d_0\alpha*\Delta T[/tex]
[tex]\Rightarrow d_f=d_0(\alpha \Delta T+1)[/tex]
So, letting d= diameter of steel rod and d'= diameter of brass ring and setting the two equal to each other I have:
[tex]d_0(\alpha \Delta T+1)=d'_0(\alpha' \Delta T+1)[/tex]
[tex]\Rightarrow d_0(\alpha \Delta T+1)-d'_0(\alpha' \Delta T+1)=0[/tex]
[tex]\Rightarrow d_0\alpha \Delta T+d_0-d'_0\alpha' \Delta T+d'_0=0[/tex]
[tex]\Rightarrow d_0\alpha \Delta T-d'_0\alpha' \Delta T=-d_0-d'_0[/tex]
[tex]\Rightarrow T_f=\frac{-d_0-d'_0}{d_0\alpha-d'_0\alpha'}+T_0[/tex]
Now d_0=3.0 cm and d'_0=2.992 cm
alpha=11*10^(-6) and alpha'= 19*10^(-6)
and T_0=25
I am getting a number like 280,000 degrees which is obviously ridiculous.
The answer is supposed to be 360 celsius. What am I screwing up here?
Thanks,
Casey
I am using rate of linear expansion: [tex]\Delta d=d_0\alpha*\Delta T[/tex]
[tex]\Rightarrow d_f=d_0(\alpha \Delta T+1)[/tex]
So, letting d= diameter of steel rod and d'= diameter of brass ring and setting the two equal to each other I have:
[tex]d_0(\alpha \Delta T+1)=d'_0(\alpha' \Delta T+1)[/tex]
[tex]\Rightarrow d_0(\alpha \Delta T+1)-d'_0(\alpha' \Delta T+1)=0[/tex]
[tex]\Rightarrow d_0\alpha \Delta T+d_0-d'_0\alpha' \Delta T+d'_0=0[/tex]
[tex]\Rightarrow d_0\alpha \Delta T-d'_0\alpha' \Delta T=-d_0-d'_0[/tex]
[tex]\Rightarrow T_f=\frac{-d_0-d'_0}{d_0\alpha-d'_0\alpha'}+T_0[/tex]
Now d_0=3.0 cm and d'_0=2.992 cm
alpha=11*10^(-6) and alpha'= 19*10^(-6)
and T_0=25
I am getting a number like 280,000 degrees which is obviously ridiculous.
The answer is supposed to be 360 celsius. What am I screwing up here?
Thanks,
Casey