Question

A rod of negligible heat capacity has length 20 cm, area of cross section $1.0c{{m}^{2}}$ and thermal conductivity $200W{{m}^{-1}}^{o}{{C}^{-1}}$. The temperature of one end is maintained at ${{0}^{0}}C$and that of the other end is slowly and linearly varied from ${{0}^{0}}C$ to ${{60}^{0}}C$ in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

Answer 1

To solve this question, we need to use the concept of heat capacity and heat transfer. The heat capacity is defined as the amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Find the rate of change of temperature and then find the heat transmitted using the discrete summation.

Complete step-by-step answer:
Given, the length of the rod is $l=20cm=0.2m$
The cross-sectional area of the rod is $A=1.0c{{m}^{2}}={{10}^{-4}}{{m}^{3}}$
Thermal conductivity k is given as, $k=200W{{m}^{-1o}}{{C}^{-1}}$
The temperature of one end of the rod is ${{0}^{0}}C$ and the temperature of the other end is linearly varies from ${{0}^{0}}C$ to ${{60}^{0}}C$ in 10 minutes.
So, the rate of change of temperature is,
$=\dfrac{\Delta T}{\Delta t}=\dfrac{60-0}{10\times 60}={{0.1}^{o}}C/s$
Now, the rate of heat transmitted through the rod can be mathematically expressed as,
$\dfrac{\Delta Q}{\Delta t}=\dfrac{\Delta TkA}{l}$
Where, $\Delta Q$ is the amount of heat transmitted, $\Delta T$ is the change in temperature, k is the thermal conductivity, A is the cross sectional area of the rod and l is the length of the rod.
So, the heat transmitted can be defined as
$Q=\sum{\dfrac{\Delta TkA}{l}\Delta t}$
Discretely, we can define the heat transferred for each small interval of time. Considering the time interval as $\Delta t=1s$, from the rate of increase of temperature, we can define that $\Delta T\Delta t$ will increase by o.1 each time.
So,
$\begin{aligned} & Q=\dfrac{kA}{l}\times 0.1+\dfrac{kA}{l}\times 0.2+\dfrac{kA}{l}\times 0.3+........+\dfrac{kA}{l}\times 60 \\\ & Q=\dfrac{kA}{l}\times \left[ 0.1+0.2+0.3+.....+60 \right] \\\ \end{aligned}$
Now, from the series summation and the sum of n terms of AP, $S=\dfrac{n}{2}(a+{{a}_{n}})$
For the series $0.1+0.2+0.3+.....+60$, the summation of the series is,
$0.1+0.2+0.3+.....+60=\dfrac{600}{2}\left( 0.1+60 \right)=300\times 60.1$
So, the heat transmitted is,
$\begin{aligned} & Q=\dfrac{200\times {{10}^{-4}}}{0.2}\times 300\times 60.1 \\\ & Q=1800J \\\ \end{aligned}$
The total heat transmitted is 1800J.

Note: Heat can be measured as the total internal energy of the system. The dependence of heat energy with temperature can be defined in terms of the physical quantities specific heat capacity and the heat capacity of the system.