Question

One end of a brass rod $2m$ long and having$1cm$radius is maintained at $250^\circ C$. When a steady state is reached, the rate of heat flow across any cross section is $0.5cal{S^{ - 1}}$. What Is the temperature of the other end $K = 0.26cal{s^{ - 1}}c{m^{ - 1}}^\circ {C^{ - 1}}.$
A. $127.6^\circ C$
B. $127.6K$
C.$122.4^\circ C$
D.$122.4K$

Answer 1

Rate of heat flow is the amount of heat transferred per unit time in some material i.e. $\dfrac{Q}{t}$
Formula used:
$\dfrac{Q}{t} = \dfrac{{K({T_1} - {T_2})A}}{{\Delta x}}$

Complete step by step answer:
Given,$\dfrac{Q}{t} = 0.5cal{s^{ - 1}}$
Value of $K = 0.26cal{s^{ - 1}}c{m^{ - 1^\circ }}{c^{ - 1}}$
The unit of $K$ is given in the CGS system. So we first need to convert the units of all the given quantities into a CGS system.
Length of rod$= 2m = 200cm$
i.e. $\Delta x = 200cm$
Radius$= 1cm$
Cross sectional area of the other end of the brass rod will be a circle.
$\therefore Area = \pi {r^2}$
$= 3.14 \times {(1)^2}$
$= 3.14c{m^2}$
We have to find the temperature of the other end i.e.${T_2}$
We know the temperature of the first end i.e. ${T_1} = 250^\circ C$
Now use formula,
$\dfrac{Q}{t} = \dfrac{{K({T_1} - {T_2})A}}{{\Delta x}}$
Where, $Q$ is the amount of heat transferred.
$t$ is time taken
$T$is temperature
$A$is area of cross section
$\Delta x$is the change in length.
${T_1} - {T_2} = \dfrac{{\Delta xQ}}{{KAt}}$
Substitute the values in the above equation.
$\Rightarrow {T_1} - {T_2} = \dfrac{{200 \times 0.5}}{{3.14 \times 0.26}}$
By simplifying the fraction, we get
${T_1} - {T_2} = \dfrac{{100}}{{0.8164}}$
$= 1.224 \times 100$
$\Rightarrow {T_1} - {T_2} = 122.4^\circ C$
But it is given that ${T_1} = 250^\circ C$
Therefore, the equation ${T_1} - {T_2} = 122.4^\circ C$
which can be rearranged as ${T_2} = {T_1} - 122.4^\circ C$
will give ${T_2} = 250^\circ C - 122.4^\circ C$
$\Rightarrow {T_2} = 127.6^\circ C$
This is the temperature at another end. So option (A) $127.6^\circ C$is correct.

Note: While solving a numerical. Make sure that all the quantities are in the same system of units, i.e. SI unit or CGS unit, to get the right answer. Even thought, temperature of one end is constant, the temperature of the other end is always less than that. This explains the concept of loss of energy. We should know that there is always a loss of energy. In reality, no system is perfect.