Question

The thermal capacities of the two bodies are in the ratio of 1:4. If the rate of loss of heat is equal for the two bodies under identical conditions of surroundings, then the ratio of the rate of fall of temperature of the two bodies is:

Answer 1

Thus the ratio of temperature of the two bodies is (dT / dt)1 / (dT / dt)2 = 4/1

Explanation:

RAte of heat transfer:

dQ / dt = ms dT / dt

The rate of loss of heat are equal for the two bodies under identical conditions of surroundings, then

(dQ / dt)1 = (dQ / dt)2

Ratio of thermal capacities.

1 x (dT / dt)1 = 4 x (dT / dt)2

(dT / dt)1 / (dT / dt)2 = 4 / 1

Thus the ratio of temperature of the two bodies is (dT / dt)1 / (dT / dt)2 = 4/1

Answer 2

Given :

Ratio of thermal capacities of the two bodies is 1 : 4 i.e $\dfrac{m_{s1}}{m_{s2}}=\dfrac{1}{4}$ ...... ( 1 ) .

To find :

The ratio of the rate of fall of temperature of the two bodies.

Solution :

We know , thermal capacity id given by :

$\dfrac{d\phi}{dt}=m_s\dfrac{d\theta}{dt}$

Here , $m_s$ is the thermal capacity .

Now , for two conditions .

$\dfrac{\dfrac{d\phi}{dt}}{\dfrac{d\phi}{dt}}=\dfrac{m_{s1}\dfrac{d\theta_1}{dt}}{m_{s2}\dfrac{d\theta_2}{dt}}\\\\\\ \dfrac{\dfrac{d\theta_1}{dt}}{\dfrac{d\theta_2}{dt}}=\dfrac{m_{s2}}{m_{s1}}$

Therefore :

$\dfrac{\dfrac{d\theta_1}{dt}}{\dfrac{d\theta_2}{dt}}=\dfrac{4}{1}$

From equation 1 .

Therefore , the rate of fall of temperature of the two bodies is 4 : 1 .

Learn More :

Thermodynamics

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