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Heat and Thermodynamics question

2021 · Shift 1 · Q57
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Heat and Thermodynamics question

2021 · Shift 1 · Q57

JEE AdvancedPhysicsHeat and ThermodynamicsNumerical+4 / −1
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 = t1 and 50 K at t = t2. 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) is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 9

  1. Use Stefan–Boltzmann law for radiative cooling

Since the object is a black body inside a large evacuated container maintained at 0 K0\,\text{K}0K, it loses energy only by radiation.

For a black body at temperature TTT, radiated power is

P=σAT4P = \sigma A T^4P=σAT4

where σ\sigmaσ is Stefan–Boltzmann constant and AAA is the surface area of the object.

Because the surroundings are at 0 K0\,\text{K}0K, there is no incoming radiation, so net power loss is

dQdt=−σAT4\frac{dQ}{dt} = -\sigma A T^4dtdQ​=−σAT4
  1. Relate heat loss to temperature change

Heat content changes as

dQ=C dTdQ = C\,dTdQ=CdT

where CCC is the heat capacity of the object, given constant.

Thus,

CdTdt=−σAT4C\frac{dT}{dt} = -\sigma A T^4CdtdT​=−σAT4

So,

dTdt=−σACT4\frac{dT}{dt} = -\frac{\sigma A}{C}T^4dtdT​=−CσA​T4
  1. Separate variables and integrate

Let

k=σACk = \frac{\sigma A}{C}k=CσA​

Then

dTdt=−kT4\frac{dT}{dt} = -kT^4dtdT​=−kT4

which gives

dTT4=−k dt\frac{dT}{T^4} = -k\,dtT4dT​=−kdt

Integrate from T=200T=200T=200 at t=0t=0t=0 to temperature TTT at time ttt:

∫200TT−4 dT=−k∫0tdt\int_{200}^{T} T^{-4}\,dT = -k\int_0^t dt∫200T​T−4dT=−k∫0t​dt

Now,

∫T−4dT=−13T3\int T^{-4}dT = -\frac{1}{3T^3}∫T−4dT=−3T31​

So,

[−13T3]200T=−kt\left[-\frac{1}{3T^3}\right]_{200}^{T} = -kt[−3T31​]200T​=−kt −13T3+13(200)3=−kt-\frac{1}{3T^3} + \frac{1}{3(200)^3} = -kt−3T31​+3(200)31​=−kt kt=13T3−13(200)3kt = \frac{1}{3T^3} - \frac{1}{3(200)^3}kt=3T31​−3(200)31​

Thus,

t=13k(1T3−12003)t = \frac{1}{3k}\left(\frac{1}{T^3} - \frac{1}{200^3}\right)t=3k1​(T31​−20031​)
  1. Find t1t_1t1​ when T=100 KT=100\,\text{K}T=100K
t1=13k(11003−12003)t_1 = \frac{1}{3k}\left(\frac{1}{100^3} - \frac{1}{200^3}\right)t1​=3k1​(10031​−20031​)

Since

12003=18⋅1003\frac{1}{200^3} = \frac{1}{8\cdot 100^3}20031​=8⋅10031​

we get

t1=13k(11003−18⋅1003)=13k⋅78⋅1003t_1 = \frac{1}{3k}\left(\frac{1}{100^3} - \frac{1}{8\cdot 100^3}\right) = \frac{1}{3k}\cdot \frac{7}{8\cdot 100^3}t1​=3k1​(10031​−8⋅10031​)=3k1​⋅8⋅10037​
  1. Find t2t_2t2​ when T=50 KT=50\,\text{K}T=50K
t2=13k(1503−12003)t_2 = \frac{1}{3k}\left(\frac{1}{50^3} - \frac{1}{200^3}\right)t2​=3k1​(5031​−20031​)

Now,

1503=81003,12003=18⋅1003\frac{1}{50^3} = \frac{8}{100^3}, \qquad \frac{1}{200^3} = \frac{1}{8\cdot 100^3}5031​=10038​,20031​=8⋅10031​

So,

t2=13k(81003−18⋅1003)=13k⋅638⋅1003t_2 = \frac{1}{3k}\left(\frac{8}{100^3} - \frac{1}{8\cdot 100^3}\right) = \frac{1}{3k}\cdot \frac{63}{8\cdot 100^3}t2​=3k1​(10038​−8⋅10031​)=3k1​⋅8⋅100363​
  1. Take the ratio
t2t1=638⋅100378⋅1003=637=9\frac{t_2}{t_1} = \frac{\frac{63}{8\cdot 100^3}}{\frac{7}{8\cdot 100^3}} = \frac{63}{7} = 9t1​t2​​=8⋅10037​8⋅100363​​=763​=9

Therefore,

t2t1=9\boxed{\frac{t_2}{t_1} = 9}t1​t2​​=9​
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