A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time , the temperature of the object is 200 K. [2021]

Question

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time $t = 0$ , the temperature of the object is 200 K. The temperature of the object becomes 100 K at $t = t_{1}$ and 50 K at $t = t_{2}$ . Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio $(\frac{t_{2}}{t_{1}})$ is _____________. [2021]

Smart Achievers · Accepted Answer

(9)

$Heat radiated = σ A T^{4} = - m s \frac{d T}{d t}$

$So,$ $\int_{200}^{100} \frac{d T}{T^{4}} = \int_{0}^{t_{1}} k d t$ $\Rightarrow \frac{1}{3 T^{3}} |_{200}^{100} = k t_{1}$

or, $\frac{1}{3} (\frac{1}{100^{3}} - \frac{1}{200^{3}}) = k t_{1}$ ...(i)

Similarly, $\frac{1}{3 T^{3}} |_{200}^{50} = k t_{2}$

$\Rightarrow \frac{1}{3} (\frac{1}{50^{3}} - \frac{1}{200^{3}}) = k t_{2}$ ...(ii)

Dividing eq. (ii) by (i),

$\frac{t_{2}}{t_{1}} = (\frac{200^{3} - 50^{3}}{200^{3} - 100^{3}}) \frac{100^{3}}{50^{3}} = 9$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (9) Heat radiated =σAT4=-msdTdt So, ∫200100dTT4=∫0t1k dt⇒13T3|200100=kt1 or, 13(11003-12003)=kt1 ...(i) Similarly, 13T3|20050=kt2 ⇒13(1503-12003)=kt2 ...(ii) Dividing eq. (ii) by (i), t2t1=(2003-5032003-1003)1003503=9

Want Us to Email you About Special Offers & Updates?