Wood in Benzene

Task number: 1805

The density of benzene is 900 kg m−3 at a temperature of 0 °C; the thermal coefficient of volume expansion is 12 · 10−4 K−1. A wooden block with a density of 880 kg m−3 floats on the surface of benzene with temperature of 0 °C. At what temperature does the body begin to sink, provided that the 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 = 0 °C  initial temperature
    ρ0b = 900 kg m−3 density of benzene at temperature t0
    βb = 12·10−4 K−1 coefficient of volume thermal expansion of benzene
    ρ0w = 880 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:

    \[t=t_0+\frac{\rho_{0\mathrm{b}}-\rho_{0\mathrm{w}}}{\rho_{0\mathrm{w}}\beta_{\mathrm{b}}-\rho_{0\mathrm{b}}\beta_{\mathrm{w}}}=\left(0+\frac{900-880}{880\cdot{12}\cdot{10^{-4}}-900\cdot{2.2}\cdot{10^{-5}}}\right){\, ^\circ}\mathrm{C}\] \[t\,\dot=\,19.3\,^{\circ}\mathrm{C}\]
  • Answer

    A wooden block starts to sink after exceeding the temperature of approximately 19.3 °C.

Difficulty level: Level 3 – Advanced upper secondary level
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