Thermodynamics #2?

Question

A glass ball is immersed into a fluid at two different temperatures t 1 =17 degrees C and t 2 = 20 degrees C. At t1 the ball displaces the mass of fluid m 1 = 30.10 g and at t 2, m 2 = 30.01 g. Calculate the difference between the coefficient of volume thermal expansion of the fluid and the corresponding coefficient for glass.
Give your answer with 3 sig. fig. in scientific notation.

Answer 1

Nice problem. There are two approaches, the trusting and the suspicious.

In the trusting approach, we assume that question makes sense and that all that counts is the difference between the coefficients. So we simplify our calculations by assuming that the coefficient for glass is 0.

Then at temperature T0 volume V0 contains M0 grams of fluid, while at temperature T1 the same volume contains M1 grams of fluid.

To get the volume change from the mass change, we go through densities:

Density = mass/volume
D0 = M0/V0
D1 = M1/V0

But:
volume = mass/density
V(T0) = V0
V(T1) = M0/D1 = M0/(M1/V0) = V0(M0/M1)

So a change in temperature from T0 to T1 causes fractional change in volume of

V(T1)/V(T0) = M0/M1

If (T1 - T0) is small enough, we can just divide to get an approximation to the coefficient of volume thermal expansion.

Otherwise, it is essentially a compound interest problem with the coefficient of expansion acting as the interest rate, the ratio acting as the final value, 1 as the principle value, and (T1 - T0) as the interval:

http://en.wikipedia.org/wiki/Compound_interest

You can go to a table or on-line app to compute the coefficient of expansion.

If you want to go with the more suspicious method, then you have to expect that the T1 volume of the glass ball is larger than the T0 volume by the appropriate coefficient so the equations get more complicated. You can try to deal with them symbolically, or you can just use a non-zero value for the glass coefficient and see what you come with numerically.

(In fact you don't have to be suspicious because trusting works, but you should think about why that is. The easiest approach is to note that if the glass ball is _defined_ as the unit volume, whatever its temperature, ...)