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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2017 · Paper 1 · Physics · Thermal Properties of Matter

A human body has a surface area of approximately 1 m ^ 2 . The...

A human body has a surface area of approximately 1m21{m}^{2} . The normal body temperature is 10K10K above the surrounding room temperature T0{T}_{0} . Take the room temperature to be T0=300K{T}_{0} = 300K . For T0=300K{T}_{0} = 300K , the value of σT4=460Wm2\sigma{T}^{4} = 460\dfrac{W}{{m}^{2}} (where σ\sigma is the Stefan-Boltzmann constant). Which of the following options is/are correct?

Options

  1. AIf the surrounding temperature reduces by a small amount {&Delta;T}_{0} < < {T}_{0} , then to maintain the same body temperature the same (living) human being needs to radiate &Delta;W = 4&sigma;{T}_{0}^{3}{&Delta;T}_{0} more energy per unit time
  2. BReducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation.
  3. CIf the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths
  4. DThe amount of energy radiated by the body in 1 second is close to 60Joules60Joules

Correct answer

C. If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

Step-by-step solution

 Energy radiated =σA(T4−T04)t=σA[(T0+10)4−T04]t=σAT04[(1+10T0)4−1]t=σAT04[40T0]×t=460×1×40300×1=61.33JP= Energy radiated  time =σAT4−σAT04∴∣dpdT0∣=σA(4T03)∴∣dp∣=σA(4T03)dT0∴∣ΔP∣=4σAT03\begin{aligned} & \text{ Energy radiated }=\sigma A\left(T^4-T_0^4\right) \mathrm{t} \\ & =\sigma A\left[\left(\mathrm{ T}_0+10\right)^4-T_0^4\right] t \\ & =\sigma A T_0^4\left[\left(1+\frac{10}{T_0}\right)^4-1\right] t \\ & =\sigma A T_0^4\left[\frac{40}{T_0}\right] \times t=460 \times 1 \times \frac{40}{300} \times 1=61.33 J \\ & P=\frac{\text{ Energy radiated }}{\text{ time }}=\sigma A T^4-\sigma A T_0^4 \\ & \therefore\left|\frac{d p}{d T_0}\right|=\sigma A\left(4 T_0^3\right) \therefore|d p|=\sigma A\left(4 T_0^3\right) d T_0 \\ & \therefore|\Delta P|=4 \sigma A T_0^3 \end{aligned}) A, B are not correct options as human body is not a black body. Energy radiated (\propto \mathrm{A}) where (\mathrm{A}) isthethesurface area of the body. ' (\mathrm{C}) ' is the correct option$
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