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Physics · Heat Transfer

JEE Advanced 2021 — Paper 1 — Question 11

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\mathrm{t}=0, the temperature of the object is 200 K . The temperature of the object becomes 100 K at t=t1\mathrm{t}=\mathrm{t}_{1} and 50 K at t=t2\mathrm{t}=\mathrm{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 (t2/t1)\left(t_{2} / t_{1}\right) is \qquad ,

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

−CdTdt=(T4−Ts4)-C \frac{d T}{d t}=\left(T^{4}-T_{s}^{4}\right) ∫200100dTT4−TS4=∫0t1−1cdt\int_{200}^{100} \frac{d T}{T^{4}-T_{S}^{4}}=\int_{0}^{t_{1}}-\frac{1}{c} d t

−13[1 T3]200100=−1C(t1)-\frac{1}{3}\left[\frac{1}{\mathrm{~T}^{3}}\right]_{200}^{100}=-\frac{1}{\mathrm{C}}\left(\mathrm{t}_{1}\right)

⇒[1(100)3−1(200)3]=3Ct1\Rightarrow\left[\frac{1}{(100)^{3}}-\frac{1}{(200)^{3}}\right]=\frac{3}{\mathrm{C}} \mathrm{t}_{1}

Similarly, [1(50)3−1(200)3]=3Ct2\left[\frac{1}{(50)^{3}}-\frac{1}{(200)^{3}}\right]=\frac{3}{\mathrm{C}} \mathrm{t}_{2}

t2t1=[1(50)3−1(200)3][1(100)3−1(200)3]=9\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{\left[\frac{1}{(50)^{3}}-\frac{1}{(200)^{3}}\right]}{\left[\frac{1}{(100)^{3}}-\frac{1}{(200)^{3}}\right]}=9

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Physics
Chapter
Heat Transfer
Topic
Convection and Radiation