Question

A metal rod of cross-sectional area $1c{{m}^{2}}$ is being heated at one end. At some instant, the temperature gradient is $5{}^\circ Cc{{m}^{-1}}$ at a cross-section $A$ and is $2.5{}^\circ Cc{{m}^{-1}}$ at cross-section $B$. The heat capacity of the part $AB$ of the rod is $0.4J{}^\circ {{C}^{-1}}$. Thermal conductivity of the material of the rod is $200W{{m}^{-1}}{}^\circ {{C}^{-1}}$. Neglect any loss of heat to the atmosphere. Calculate the rate at which the temperature is increasing in the part $AB$ of the rod.
A) $24{}^\circ C{{s}^{-1}}$
B) $40{}^\circ C{{s}^{-1}}$
C) $12.5{}^\circ C{{s}^{-1}}$
D) $48{}^\circ C{{s}^{-1}}$

Answer 1

The given temperature gradients at two instants are used to find the rate of flow of heat at those particular instants. From the expression relating specific heat capacity and rate of heat flow of a conducting material, we can derive the rate of change of temperature within these two instants.

Complete answer:
Let us consider a metal rod of cross-sectional area $1c{{m}^{2}}$. One end of the rod is heated. It is obvious that the other end of the rod will get heated in some time. We are supposed to consider two instants of time or two cross- sectional areas $A$ and $B$ through which the heat flows. It is given that the temperature gradients on these cross-sectional areas are $5{}^\circ Cc{{m}^{-1}}$ and $2.5{}^\circ Cc{{m}^{-1}}$, respectively. Temperature gradient is a physical quantity which describes the direction of flow of temperature as well as the rate of flow of temperature through a particular location. It is expressed as $\dfrac{dT}{dx}$when the flow of temperature is in the $x$ direction. Let us express the temperature gradient at the cross-section $A$ as ${{\left( \dfrac{dT}{dx} \right)}_{A}}$ and the temperature gradient at the cross-section $B$ as ${{\left( \dfrac{dT}{dx} \right)}_{B}}$. Clearly,

${{\left( \dfrac{dT}{dx} \right)}_{A}}=5{}^\circ Cc{{m}^{-1}}=500{}^\circ C{{m}^{-1}}$
Let this be equation 1.
Also,
${{\left( \dfrac{dT}{dx} \right)}_{B}}=2.5{}^\circ Cc{{m}^{-1}}=250{}^\circ C{{m}^{-1}}$
Let this be equation2.

Now, let us find the rate of flow of heat along $AB$. We know that rate of flow of heat is expressed as
$\dfrac{d\theta }{dt}=KA\dfrac{dT}{dx}$
Where
$\dfrac{d\theta }{dt}$ is the rate of flow of heat along a section of a heat conducting material
$K$ is the thermal conductivity of the heat conducting material
$A$ is the area of the heat conducting material
$\dfrac{dT}{dx}$ is the temperature gradient at an instant of time, provided the direction of heat flow is in the $x$ direction.
Let this be equation 1.

We are provided that the thermal conductivity of the metal rod is
$K=200W{{m}^{-1}}{}^\circ {{C}^{-1}}$
We are also given that the area of cross-section of the metal rod is
$a=1c{{m}^{2}}=0.0001{{m}^{2}}$
Let the heat flow along the cross-sectional area $A$ be ${{\left( \dfrac{d\theta }{dt} \right)}_{A}}$ and the heat flow along the cross-sectional area $B$ be ${{\left( \dfrac{d\theta }{dt} \right)}_{B}}$. Clearly, from equation 3 and equation 1, we have

${{\left( \dfrac{d\theta }{dt} \right)}_{A}}=Ka{{\left( \dfrac{dT}{dx} \right)}_{A}}=Ka(500)=200\times 0.0001\times 500=10J{{s}^{-1}}$
Let this be equation 4.

Also, from equation 3 and equation 2, we have
${{\left( \dfrac{d\theta }{dt} \right)}_{B}}=Ka{{\left( \dfrac{dT}{dx} \right)}_{B}}=Ka(250)=200\times 0.0001\times 250=5J{{s}^{-1}}$
Let this be equation 5.

Now, the rate of flow of heat through $AB$ is given by
${{\left( \dfrac{d\theta }{dt} \right)}_{AB}}={{\left( \dfrac{d\theta }{dt} \right)}_{A}}-{{\left( \dfrac{d\theta }{dt} \right)}_{B}}=10-5=5J{{s}^{-1}}$
Let this be equation 6.

We know the general expression that the change in heat in a conducting material is equal to the product of the heat capacity of the material and the change in temperature.
$d\theta =CdT$
where
$d\theta$ is the change in heat
$C$ is the heat capacity of the material
$dT$ is the change in temperature
To get the rate of change in heat in the material, let us divide the above equation by $dt$ as follows.
$\dfrac{d\theta }{dt}=C\dfrac{dT}{dt}$
where
$\dfrac{d\theta }{dt}$ is the rate of heat flow
$C$ is the heat capacity of the material
$\dfrac{dT}{dt}$ is the rate of change of temperature
Let this be equation 7.

Substituting equation 6 in equation 7, we have
${{\left( \dfrac{d\theta }{dt} \right)}_{AB}}=C{{\left( \dfrac{dT}{dt} \right)}_{AB}}$
where
${{\left( \dfrac{d\theta }{dt} \right)}_{AB}}$ is the rate of heat flow in the section $AB$ of the metal rod
${{\left( \dfrac{dT}{dt} \right)}_{AB}}$ is the rate of temperature change in the section $AB$ of the metal rod

Rearranging the above expression,
${{\left( \dfrac{dT}{dt} \right)}_{AB}}=\dfrac{{{\left( \dfrac{d\theta }{dt} \right)}_{AB}}}{C}=\dfrac{5}{0.4}=12.5{}^\circ C{{s}^{-1}}$
Therefore, the rate at which temperature is increasing in the part $AB$ of the rod is equal to $12.5{}^\circ C{{s}^{-1}}$.

So, the correct answer is “Option C”.

Note:
It is important to convert the units of the quantities given in the question to the same unit system. It is always better to convert the values to the International System of Units (SI) for easy calculations. Students need to be thoroughly aware of the SI units of various quantities as well as the related conversions for easy mathematics.