# Homework Help: Thermal expansion of a liquid in a cylindrical container

1. Mar 21, 2016

### issacnewton

1. The problem statement, all variables and given/known data
There is liquid in a cylindrical container at some level. Now when the container is heated, the level of the liquid remains the same in the container. What is the relationship between the coefficient of linear thermal expansion of liquid and the container ?

2. Relevant equations
$$\Delta V = V_o 3\alpha \Delta T$$
$$\Delta l = l_o \alpha \Delta T$$

3. The attempt at a solution

Now as the temperature increases, all dimensions of the container increase. Its height increases and also its radius increases. Since liquid remains at the same level, does that mean that the change in volume for both of them is the same ? Please guide... If we mark the level of the liquid before heating by some marker, then the liquid remains at this marker position even after the heating.

Thanks

2. Mar 21, 2016

### BvU

What is the relevant equation for the liquid level h (as a function of V) ?

3. Mar 21, 2016

### issacnewton

I think I found the answer...... Let original radius of cylinder be r and original liquid marker position be x. Then the original volume of the liquid is
$$\pi r^2 x$$
Let new radius of the container be $r'$ and new liquid marker position be $x'$. Let $\alpha_c$ be the coefficient of the linear thermal expansion of the container and let $\Delta T$ be the change in temperature.. Then we have
$$r' = r [ 1 + \alpha_c \Delta T]$$
$$x' = x [ 1 + \alpha_c \Delta T]$$

Now the new volume of the liquid is $\pi r'^2 x'$. Which is equal to
$$\pi r^2 [ 1 + \alpha_c(\Delta T)]^2 x [ 1 + \alpha_c(\Delta T)]$$
$$\pi r^2 x [ 1 + \alpha_c(\Delta T)]^3$$

So we have
$$\frac{V^{liquid}_{new}}{V^{liquid}_{old}} = [ 1 + \alpha_c(\Delta T)]^3$$

But the new volume of liquid is related to old volume of liquid by the relation involving the coefficient of volume thermal expansion of the liquid.

$$V^{liquid}_{new} = V^{liquid}_{old} [ 1 + \beta_{liq} (\Delta T)]$$

Comparing two equations, we get

$$1 + \beta_{liq} (\Delta T) = [ 1 + \alpha_c(\Delta T)]^3$$

Since $\alpha_c$ is very small for most of the materials, we can drop higher order terms here.

$$1 + \beta_{liq} (\Delta T) \approx 1 + 3 \alpha_c(\Delta T)$$

Would this be correct ?