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.
