Question

Spheres P and Q are uniformly constructed from the same material which is a good conductor of heat and the radius of Q is thrice the radius of P. The rate of fall temperature of P is x times that of Q. When both are at the same surface temperature, the value of x is:
A. $\dfrac{1}{4} $
B. $\dfrac{1}{3} $
C. 3
D. 4

Answer 1

Hint: If both spheres are heated to a temperature T, then we need to find the equation for rate of heat loss with respect to the radius as the difference between the two spheres is just their radius. We are also given that the rate of temperature change of Q is x times of P.

Formula used:
The rate of change of temperature or (or fall) is given as:
$\dfrac{\Delta T}{\Delta t} = \dfrac{e \sigma A T^4}{m C}$,
where A is the cross section area, T is temperature, C is specific heat, e is emissive power of the body, $\sigma$ is Stefan's constant.

Complete answer:
The two spheres cool down by giving out radiation. Rate of loss of heat due to radiation is given as:
$\dfrac{\Delta Q}{\Delta t} = e \sigma A T^4$
Also, we are familiar with another expression:
$\Delta Q = m C \Delta T$.
On combining the two expressions we get:
$\dfrac{\Delta T}{\Delta t} = \dfrac{e \sigma A T^4}{m C}$ ;
here,
$A = \pi r^2$
$m = \rho \times (4/3) \pi r^3$
which gives us $r^2$ in numerator and $r^3$ in denominator. So,
$\dfrac{\Delta T}{\Delta t} \propto \dfrac{1}{r}$
As the other quantities will be same for the two spheres, we may write:
$\dfrac{ \left( \dfrac{\Delta T}{\Delta t} \right)_Q}{\left( \dfrac{\Delta T}{\Delta t} \right)_P} = \dfrac{r_P}{r_Q}$ .
Now, we are given that
$r_Q = 3 r_P$
So,
$\dfrac{ \left( \dfrac{\Delta T}{\Delta t} \right)_Q}{\left( \dfrac{\Delta T}{\Delta t} \right)_P} = \dfrac{1}{3}$
We are also given that rate of fall of temperature of P is x times that of Q or,
$\left( \dfrac{\Delta T}{\Delta t} \right)_P = x \left( \dfrac{\Delta T}{\Delta t} \right)_Q$
Therefore, x = 3 comes out as a conclusion if we nearly compare this with the result obtained.

So, the correct answer is option (C).

Note:
Option B and option C can be confusing. If you just make a mistake in interpretation of the statement of the question you might conclude the wrong answer. The formula for temperature variation rate is obtained as a result of black body radiation Stefan's law. There are different laws for black body radiation so one must be careful.