The separation between atoms and molecules increases when temperature rises. This leads to thermal expansion of the material. The thermal expansion of solid objects due to temperature change $\Delta T$ is characterized by
The fractional changes in length, area and volume are proportional to the change in temperature. The coefficient of linear, areal and volumetric expansion are related by $\beta=2\alpha$ and $\gamma=3\alpha$ i.e., \begin{align} &\alpha:\beta:\gamma=1:2:3 \end{align}
In general, objects will expand with increasing temperature. However, water expands with increasing temperature when it is at temperatures greater than 4℃ but it contracts when it is between 4℃ and 0℃ (water is densest at 4℃).
The thermal expansion of liquids (at pressure $p$) is characterized by the coefficient of volumetric expansion defined as \begin{align} \gamma=\left.\frac{1}{V}\frac{\mathrm{d} V}{\mathrm{d} T}\right|_{p}\nonumber \end{align}
In gases, the pressure $p$ may be constant (isobaric process) or varying (adiabatic process). The thermal expansion of gases is characterized by the coefficient of volumetric expansion defined for the particular process ( see IIT JEE 2008 problem ).
The temperatures rise leads to thermal stresses if an object is not allowed to expand e.g., a metal bar fixed at both ends. These stresses are calculated by equating increase in length (thermal expansion) and decrease in length (strain due to mechanical stress).
An ideal gas is expanding such that $PT^2=\text{constant}$. The coefficient of volume expansion of the gas is,
Solution: The coefficient of volume expansion is defined as, \begin{align} \label{bga:eqn:1} & \gamma=\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}T}. \end{align} The ideal gas equation, $PV=nRT$, gives $P=nRT/V$. Substitute $P$ in $PT^2=\text{constant}$ to get, \begin{align} \label{bga:eqn:2} & V=kT^3, \end{align} for some constant $k$. Differentiate second equation with respect to $T$ and substitute in first equation to get $\gamma=3/T$.
A cube of coefficient of linear expansion $\alpha_s$ is floating in a bath containing a liquid of coefficient of volume expansion $\gamma_l$. When the temperature is raised by $\Delta T$, the depth up to which the cube is submerged in the liquid remains the same. Find the relation between $\alpha_s$ and $\gamma_l$ showing all the steps.
Solution: Let cube has material density $\rho_s$, edge length $a$ and its height $h$ be inside the liquid of density $\rho_l$. The forces acting on the cube are its weight $a^3\rho_s g$ and the buoyancy force $a^2h\rho_l g$. In floating condition, \begin{align} \label{tsa:eqn:1} &a^3\rho_s g=a^2 h\rho_l g. \end{align} When temperature is increased by $\Delta T$, edge length of the cube increases to $a^\prime$ and density of the liquid decreases to $\rho_l^\prime$, given by, \begin{align} \label{tsa:eqn:2} a^\prime&=a(1+\alpha_s\Delta T),\\ \label{tsa:eqn:3} \rho_l^\prime&={\rho_l}/{(1+\gamma_l\Delta T)}. \end{align} Now, the volume of displaced liquid is $a^{\prime\,2}h$ and buoyancy force is $a^{\prime\,2}h\rho_l^\prime g$. Equate weight of the cube to the buoyancy force to get, \begin{align} \label{tsa:eqn:4} a^3\rho_s g&=a^{\prime\,2}hg \rho_l^\prime \nonumber\\ & =\frac{a^2(1+\alpha_s\Delta T)^2 hg \rho_l}{1+\gamma_l\Delta T}, \end{align} where we used second and third equations to eliminate $a^\prime$ and $\rho_l^\prime$. Divide first equation by fourth equation to get, \begin{align} 1+\gamma_l\Delta T&=(1+\alpha_s\Delta T)^2 \nonumber\\ &\approx 1+2\alpha_s \Delta T \nonumber \end{align} i.e., $\gamma_l=2\alpha_s$.Two rods, one of aluminium and the other made of steel, having initial lengths $l_1$ and $l_2$ are connected together to form a single rod of length $l_1+l_2$. The coefficients of linear expansion for aluminium and steel are $\alpha_a$ and $\alpha_s$, respectively. If the length of each rod increases by the same amount when their temperature are raised by $t\;\mathrm{{}^oC}$, then find the ratio ${l_1}/{(l_1+l_2)}$,
Solution: After thermal expansion, length of the aluminium and the steel rod become, \begin{align} & l_1^\prime=l_1(1+\alpha_at),\\ & l_2^\prime=l_2(1+\alpha_st). \end{align} Since length of each rod is increased by the same amount, we get, \begin{align} &l_1^\prime-l_1=l_2^\prime-l_2. \end{align} Substitute $l_1^\prime$ and $l_2^\prime$ from first and second equation into third equation to get, $l_1\alpha_a=l_2\alpha_s$. Hence, \begin{align} \frac{l_1}{l_1+l_2}=\frac{\alpha_s}{\alpha_a+\alpha_s}. \nonumber \end{align}
A bimetallic strip is formed out of two identical strips - one of copper and the other of brass. The coefficients of linear expansion of the two metals are $\alpha_C$ and $\alpha_B$. On heating, the temperature of the strip goes up by $\Delta T$ and the strip bends to form an arc of radius of curvature $R$. Then, $R$ is,
Solution: Let $d$ and $l_0$ be width and initial length of the two strips. The length of the strips after heating becomes, \begin{align} l_C&=l_0(1+\alpha_C\Delta T), \nonumber\\ l_B&=l_0(1+\alpha_B\Delta T).\nonumber \end{align}
The strips bend as shown in the figure. From figure, the length of the two strips are \begin{align} l_C&=R\theta, \nonumber\\ l_B&=(R+d)\theta. \nonumber \end{align} Divide to get, \begin{align} \label{jqa:eqn:1} \frac{R+d}{R}&=\frac{l_B}{l_C}=\frac{1+\alpha_B\Delta T}{1+\alpha_C\Delta T}\nonumber\\ &=(1+\alpha_B\Delta T)(1+\alpha_C\Delta T)^{-1}\nonumber\\ &\approx (1+\alpha_B\Delta T)(1-\alpha_C\Delta T) \nonumber\\ &\approx 1+(\alpha_B-\alpha_C)\Delta T, \end{align} where we neglected higher order terms. Simplify above equation to get, \begin{align} \label{jga:eqn:2} R=\frac{d}{(\alpha_B-\alpha_C)\Delta T}. \end{align} The bending shown in figure assume $\alpha_B > \alpha_C$ (which is generally the case for brass and copper). If $\alpha_B < \alpha_C$, bending of strip will be in opposite direction. To take care of both cases, above equation is written as, \begin{align} R=\frac{d}{|\alpha_B-\alpha_C|\,\Delta T}.\nonumber \end{align}
If a cylinder of diameter 1.0 cm at 30℃ is to be inserted into a hole of diameter 0.9997 cm in a steel plate at the same temperature, then minimum required rise in the temperature of the plate is (Coefficient of linear expansion of steel $=13 \times{10}^{-6}$/℃).
Hint: The diameter of the hole increases when the plate is heated. Substitute values in $D=D_0(1+\alpha \Delta T)$ to get $\Delta T=23$℃.
A 200 liter steel water tank is completely filled. A vertical overflow pipe of diameter 2.0 cm rises vertically from the top of tank. How high does the water rise in the overflow pipe if the water is heated from 15℃ to 40℃? (Coefficient of linear expansion of steel $=13 \times{10}^{-6}$ /℃).
A glass of water with volume 1 liter is completely filled at 5℃. How much water will spill out of the pyrax glass when the temperature is raised to 85℃? (The coefficient of volume expansion of water is $2.1\times10^{-4}$/℃ and that of pyrax is $0.09\times10^{-4}$℃)
Hint: Consider the thermal expansion of the liquid as well as glass. $\Delta V_w = 0.0168$ litre and $\Delta V_g = 0.00072$ litre. We encourage you to find general expession for apparent expansion of liquid in a container.
Question 1: The coefficient of volumetric expansion of a material of volume $V$ and density $\rho$ is given by
Question 2: A metal ball, completely immersed in alcohol, is weighed by suspending it from a spring balance. It weighs $W_1$ at 0℃ and $W_2$ at 150℃. If the coefficient of cubical expansion of the metal is less than that of alcohol and the density of metal is high as compared to that of alcohol then
Question 3: A sphere of mass $m$ and diameter $D$ is heated by temperature $\Delta T$. If the coefficient of linear expansion is $\alpha$ then change in surface area of the sphere is
Question 4: The coefficient of linear expansion of an inhomogeneous rod changes linearly from $\alpha_1$ to $\alpha_2$ from one end to the other end of the rod. The effective coefficient of linear expansion of the rod is
Question 5: The second's pendulum of a clock has a string of steel. The clock shows correct time at room temperature of 20℃. If room temperature is increased to 30℃ then (Coefficient of linear expansion of steel $=13 \times{10}^{-6}$ /℃)
Question 6: A steel meter scale of least count 1 mm is calibrated at 15℃. This scale is used to measure the length of a 60.0 cm long stick. If measurement is carried out at 27℃ then the scale will read (Coefficient of linear expansion of steel $=1.2 \times{10}^{-5}$ /℃)