Change of Volume of a Body When Being Heated

Task number: 1808

An edge length of an iron cube at a temperature of 0 °C is equal to 0.2 m. A zinc cuboid has two edges also 0.2 m long and the third edge measures 0.199 m at the same temperature.

Determine the temperature, at which both bodies have the same volume.

  • Hint

    What happens with volume of a body when we increase its temperature? What relationship describes this phenomenon?

  • Analysis

    When increasing the temperature of iron cube and zinc cuboid, the volumes of both bodies increase as a result of thermal volume expansion.

    As zinc has a greater coefficient of linear expansion than iron, the volume of zinc cuboid therefore increases more rapidly than the volume of the iron cube. The two volumes are thus at a certain temperature equal. The value of this temperature is obtained from the equality of finite volumes of both bodies and the relation of volume expansion.

  • Given values

    t1 = 0 °C initial temperature of both bodies
    a = 0.2 m edge length of iron cube and two edges of zinc cuboid
    b = 0,199 m length of third edge of zinc cuboid
    t = ? temperature at which both bodies have the same volume

    From The Handbook of Chemistry and Physics :

    αFe = 1.2·10−5 K−1 coefficient of linear expansion of iron
    αZn = 2.9·10−5 K−1 coefficient of linear expansion of zinc
  • Solution

    When increasing temperature, the edge length of an iron body increases more slowly than the edge length of a zinc body, because the expansion coefficient of iron is smaller than that of zinc. Consequently, the volumes of the two bodies will be the same at a certain temperature t > 0. To find this temperature, we first need to determine the coefficients of volume expansion of both bodies.

    The relation between the coefficient of thermal expansion α and volume expansion β is:

    \[\beta_{\mathrm{Fe}} = 3\alpha_{\mathrm{Fe}}\,,\] \[\beta_{\mathrm{Zn}} = 3\alpha_{\mathrm{Zn}}\,.\]

    We start from the equality of finite volumes of both bodies:

    \[V_{\mathrm{Fe}}=V_{\mathrm{Zn}}\]

    and substitute the relation for volume expansion:

    \[V_{\mathrm{0Fe}}\left[1+\beta_{\mathrm{Fe}}(t-t_{\mathrm{1}})\right]=V_{\mathrm{0Zn}}\left[1+\beta_{\mathrm{Zn}}(t-t_{\mathrm{1}})\right],\]

    where t1 is the initial temperature of both bodies.

    Next, we substitute the initial volumes V0Fe = a3 and V0Zn = a2b, where a, b indicate the dimensions of the bodies.

    \[a^{3}\left[1+\beta_{\mathrm{Fe}}(t-t_{\mathrm{1}})\right]=a^{2}b\left[1+\beta_{\mathrm{Z}n}(t-t_{\mathrm{1}})\right]\]

    Now we evaluate the unknown temperature t:

    \[a^{3}-a^{2}b=(t-t_{\mathrm{1}})(a^{2}b\beta_{\mathrm{Zn}}-a^{3}\beta_{\mathrm{Fe}})\,,\] \[t-t_{\mathrm{1}}=\frac{a^{3}-a^{2}b}{a^{2}b\beta_{\mathrm{Zn}}-a^{3}\beta_{\mathrm{Fe}}}\,,\] \[t=t_{\mathrm{1}}+\frac{a^{3}-a^{2}b}{a^{2}b\beta_{\mathrm{Zn}}}-a^{3}\beta_{\mathrm{Fe}}\,.\]
  • Numerical solution

    \[\beta_{\mathrm{Fe}}=3\alpha_{\mathrm{Fe}} = 3\cdot{1{.}2}\cdot{10^{-5}}\,\mathrm{K^{-1}}=3{.}6\cdot{10^{-5}}\,\mathrm{K^{-1}}\] \[\beta_{\mathrm{Zn}}=3\alpha_{\mathrm{Zn}} = 3\cdot{2{.}9}\cdot{10^{-5}}\,\mathrm{K^{-1}}=8{.}7\cdot{10^{-5}}\,\mathrm{K^{-1}}\]

     

    \[t=t_{\mathrm{1}}+\frac{a^{3}-a^{2}b}{(a^{2}b\beta_{\mathrm{Zn}}-a^{3}\beta_{\mathrm{Fe}})}=\] \[= \frac{0{.}2^{3}-0{.}2^{2}\cdot{0{.}199}}{(0{.}2^{2}\cdot0{.}199\cdot{8{.}7}\cdot{10^{-5}}-0.2^{3}\cdot{3{.}6}\cdot{10^{-5}})}\,\mathrm{^{\circ}C}\]

     

    \[t \dot= 98{.}9\,\mathrm{^{\circ}C}\]
  • Answer

    Volumes of both bodies are the same at a temperature of 98.9 °C.

Difficulty level: Level 2 – Upper secondary level
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