## Wood in Benzene

The density of benzene is 880 kg m−3 at a temperature of 10 °C; the thermal coefficient of volume expansion is 12 · 10−4 K−1. A wooden block with a density of 860 kg m−3 floats on the surface of benzene with temperature of 10 °C. At what temperature does the body begin to sink, provided that the average thermal coefficient of volume expansion of wood is 2.2 · 10−5 K−1?

• #### Hint

What is the relationship between the density of benzene and the density of wood when the block starts to sink?

• #### Analysis

The thermal coefficient of volume expansion of benzene is greater than the coefficient of wood. This means that the density of benzene decreases with increasing temperature more rapidly than the density of wood. At some point, both densities match up and then the wooden block starts to sink since its density becomes greater than the density of benzene. We want to find exactly the temperature at which both densities match up.

This temperature can be found by expressing both densities using the definition of density, the relationship for volume thermal expansion and evaluating the unknown temperature from the equality of both densities.

• #### Given values

 t0 = 10 °C initial temperature ρ0b = 880 kg m−3 density of benzene at temperature t0 βb = 12·10−4 K−1 coefficient of volume thermal expansion of benzene ρ0w = 860 kg m−3 density of wood at temperature t0 βw = 2.2·10−5 K−1 coefficient of volume thermal expansion of wood t = ? temperature at which wooden block begins to sink
• #### Solution

The wooden block begins to sink when its density is greater than the density of benzene. Our first task is to find the relation describing the dependence of the density of wood and benzene at a temperature difference Δt. We start from the definition of density ρ.

$\rho=\frac{m}{V}.$

We consider that the mass m does not change with changing temperature, and express the change of volume V using the formula for the volume expansion

$V=V_0(1+\beta\mathrm{\Delta}t).$

Therefore it is true that

$\rho_{\mathrm{b}}=\frac{m_{\mathrm{b}}}{V_{\mathrm{b}}}=\frac{m_{\mathrm{b}}}{V_{0\mathrm{b}}\left(1+\beta_{\mathrm{b}}\mathrm{\Delta}t\right)}=\frac{\rho_{0\mathrm{b}}}{1+\beta_{\mathrm{b}}\mathrm{\Delta}t}\,,$ $\rho_{\mathrm{w}}=\frac{m_{\mathrm{w}}}{V_{\mathrm{w}}}=\frac{m_{\mathrm{w}}}{V_{0\mathrm{w}}\left(1+\beta_{\mathrm{w}}\mathrm{\Delta}t\right)}=\frac{\rho_{0\mathrm{w}}}{1+\beta_{\mathrm{w}}\mathrm{\Delta}t}\,.$

Now we need to find the temperature difference Δt, at which the density of benzene ρb and wood ρw are equal

$\rho_b = \rho_w.$

After substituting from the above equations we obtain

$\frac{\rho_{0\mathrm{b}}}{1+\beta_{\mathrm{b}}\mathrm{\Delta}t}=\frac{\rho_{0\mathrm{w}}}{1+\beta_{\mathrm{w}}\mathrm{\Delta}t}$

From this equation we evaluate the temperature difference Δt

$\rho_{0\mathrm{b}}\left(1+\beta_{\mathrm{w}}\mathrm{\Delta}t\right)=\rho_{0\mathrm{w}}\left(1+\beta_{\mathrm{b}}\mathrm{\Delta}t\right)\,,$ $\rho_{0\mathrm{b}}-\rho_{0\mathrm{w}}=\mathrm{\Delta}t\left(\rho_{0\mathrm{w}}\beta_{\mathrm{b}}-\rho_{0\mathrm{b}}\beta_{\mathrm{w}}\right)\,,$ $\mathrm{\Delta}t=\frac{\rho_{0\mathrm{b}}-\rho_{0\mathrm{w}}}{\rho_{0\mathrm{w}}\beta_{\mathrm{b}}-\rho_{0\mathrm{b}}\beta_{\mathrm{w}}}\,.$

The wooden block thus begins to sink at a temperature t = t0 + Δt.

Numerical solution:

$t=t_0+\frac{\rho_{0\mathrm{b}}-\rho_{0\mathrm{w}}}{\rho_{0\mathrm{w}}\beta_{\mathrm{b}}-\rho_{0\mathrm{b}}\beta_{\mathrm{w}}}=\left(10+\frac{880-860}{860\cdot{12}\cdot{10^{-4}}-880\cdot{2.2}\cdot{10^{-5}}}\right){\, ^\circ}\mathrm{C}$ $t\,\dot=\,30\,^{\circ}\mathrm{C}$