## 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

*t*_{1}= 0 °Cinitial temperature of both bodies *a*= 0.2 medge length of iron cube and two edges of zinc cuboid *b*= 0,199 mlength of third edge of zinc cuboid *t*= ?temperature at which both bodies have the same volume *From The Handbook of Chemistry and Physics :**α*_{Fe}= 1.2·10^{−5}K^{−1}coefficient of linear expansion of iron *α*_{Zn}= 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

\[\beta_{\mathrm{Fe}} = 3\alpha_{\mathrm{Fe}}\,,\] \[\beta_{\mathrm{Zn}} = 3\alpha_{\mathrm{Zn}}\,.\]*α*and volume expansion*β*is: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

*t*_{1}is the initial temperature of both bodies.Next, we substitute the initial volumes

\[a^{3}\left[1+\beta_{\mathrm{Fe}}(t-t_{\mathrm{1}})\right]=a^{2}b\left[1+\beta_{\mathrm{Z}n}(t-t_{\mathrm{1}})\right]\]*V*_{0Fe}=*a*^{3}and*V*_{0Zn}=*a*^{2}*b*, where*a, b*indicate the dimensions of the bodies.Now we evaluate the unknown temperature

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

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

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