Question

thermodynamics

Answer 1

A steel is 3000cm in diameter at 25 *C.A brass ring has an interior diameter of 2.992 cm at 25 *C.At what common temperature will the ring just slide on the rod?

Answer 2

\[\alpha\]steel=12\[\times\]10^-6 K^-1 and alpha brass=18*10^-6K^-1

Answer 3

@Vincent-Lyon.Fr

Answer 4

shivam any idea?

Answer 5

I think instead we should be using \[\large \frac{(\Delta)L}{L}= \alpha (\Delta T)\]
Delta A -->change in volume
A-->Original Volume
alpha-->Coeffecient of linear expansion (
Delta T)-->Change in Temperature

Where your Final length should be equal to circumference of the steel rod

Answer 6

I mean the final circumference of the ring

Answer 7

@Aditya790 @Vincent-Lyon.Fr can check this. I am not certain about this :D

Answer 8

Before we begin, a small note you'd do good to remember. The coefficient of expansion is very similar to velocity. If the velocity is 5 m/s, the car (or whatever) covers 5 m distance every second. Similarly, if alpha if .005 /K, it means a material of 1 m (or cm or km, whatever) expands .005 m (or cm or km, the same thing always, obviously) when you heat it by 1 K. So, if you heated it up by 3 K, it would be 3 times 0.005 m. If you heated it up by T kelvin, it would be T times alpha. Now, if you had a 3 m rod instead of one meter, it is like 3 one meter rods. So, the expansion per kelvin would be 3 times the one meter rod.
Moral of story: expansion=alpha* temperature change*original length

I know many students memorize that \[\Delta L=\alpha L \Delta T\]
But understanding it should make thermal expansion much easier to you. I do hope you take some time out to understand it, at least, since I took the time out to type it down for you so nicely. Don't try to just apply the formulas and finish the question. You'll be wasting you time. See that you learn what you are doing in the end.

In the beginning, the brass rod is smaller in diameter. It won't fit on the steel rod. This is the case at 25 degrees.
Then you heat both the rods. Brass expands more than steel for the same temperature change. So, as you keep heating, even though brass is originally smaller in diameter, it will eventually catch up.
See if you can't use this analogy here.
There are two really slow moving cars, one behind the other.
The brass car is 0.008 cm behind the steel one. The steel car moves at 12x10^-6cm/sec (we put seconds here instead of temperature) and the brass car moves at 18x10^-6 cm/sec.
25 seconds have already passed (the original temperature was 25 *C, if you'll recall). At what time does the brass car catch up with the steel one?
It'll be easier for you to work with cars and speeds since you do it daily. Then go back to heat and try it by that approach.

Answer 9

A true ranchor chanchad das from 3 idiots @Aditya790 :D The best explanation I have come across yet in this topic :D I wish I could hand over three medals to you for this

Answer 10

y a superb @Aditya790 are u a student or a teacher?

Answer 11

A student. Finished Class XII.

Answer 12

I am not familiar with these dilation subjects.
My view is new diameter D' relative to initial one D is:
\[D'=D\:\left( 1+\alpha _{S}.\Delta T \right)\]provided α is a linear coefficient, not a volumetric one. If α is volumetric, you have to divide it by 3.

I find an increase in temperature ΔT = 448°C, so T' = 473°C

Answer 13

i gt 274.43 as the common temperature

Answer 14

and the method aditya suggested is too good

Answer 15

Post your calculations and I'll try to check them. I might have made a mistake.

Answer 16

@AravindG "and the method @aditya suggested is too good"
Agree 100% !

Answer 17

hm