A human body has a surface area of approximately 1\,m^2. The normal body temperature is 10K above the surrounding room temperature T_0. Take the room temperature to be T_0 = 300K. For T_0 = 300\,K, the value of \sigma T_0^4 = 460\,Wm^ - 2 (where \sigma is the Stefan-Boltzmann constant). Which of the following options is / are correct?

The amount of energy radiated by the body in 1 second is close to 60 joules, If the surrounding temperature reduces by a small amount \Delta T_0 \lt \lt 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, Reducing the exposed surface area of the body (e.g. by curling up) allows human to maintain the same body temperature while reducing the energy lost by radiation

The amount of energy radiated by the body in 1 second is close to 60 joules

If the surrounding temperature reduces by a small amount \Delta T_0 \lt \lt 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

Reducing the exposed surface area of the body (e.g. by curling up) allows human to maintain the same body temperature while reducing the energy lost by radiation

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

Properties of Matter PYQ — JEE Advanced 2017 · Shift 1 · Q40 with Solution

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Correct answer: A, B, C

Use Stefan–Boltzmann law for net radiative loss

A body at temperature $T$ in surroundings at temperature $T_0$ loses energy by radiation at the net rate

$P = \sigma A\left(T^4 - T_0^4\right).$

Here,

$A \approx 1\,\text{m}^2$
$T_0 = 300\,\text{K}$
body temperature $T = 310\,\text{K}$
given $\sigma T_0^4 = 460\,\text{W m}^{-2}$

So we evaluate each option.

Option A

We need the energy radiated in $1\,$ s, i.e. net power in watts.

$P = \sigma A(310^4 - 300^4).$

Since the temperature difference is small, use linearization:

$T^4 - T_0^4 \approx 4T_0^3(T-T_0).$

Thus,

$P \approx 4\sigma A T_0^3(10).$

Now from

$\sigma T_0^4 = 460,$

we get

$\sigma T_0^3 = \frac{460}{300}.$

Hence

= \frac{18400}{300} \approx 61.3\,\text{W}.$$ So in $1\,$s, energy radiated is about $$61\,\text{J},$$ which is close to $60\,$J. So **A is correct**. --- 3. **Option B** Initially, $$P = \sigma A(T^4 - T_0^4).$$ If surrounding temperature decreases by a small amount $\Delta T_0$, with body temperature kept same, then the new loss becomes $$P' = \sigma A\left(T^4 - (T_0-\Delta T_0)^4\right).$$ Increase in radiative loss: $$\Delta W = P' - P = \sigma A\left[T_0^4 - (T_0-\Delta T_0)^4\right].$$ For small $\Delta T_0$, $$(T_0-\Delta T_0)^4 \approx T_0^4 - 4T_0^3\Delta T_0.$$ Therefore, $$\Delta W \approx 4\sigma A T_0^3\Delta T_0.$$ Since $A = 1\,\text{m}^2$ here, $$\Delta W = 4\sigma T_0^3\Delta T_0.$$ So **B is correct**. --- 4. **Option C** Net radiative loss is $$P = \sigma A(T^4 - T_0^4).$$ At fixed body temperature $T$ and fixed surroundings $T_0$, the loss is directly proportional to exposed area $A$. So if a person curls up and reduces exposed surface area, then radiative energy loss decreases while body temperature can be maintained more easily. So **C is correct**. --- 5. **Option D** By Wien's displacement law, $$\lambda_{\max} T = \text{constant}.$$ If body temperature rises, then $T$ increases, so $$\lambda_{\max} \propto \frac{1}{T}$$ decreases. Thus the peak shifts to **shorter** wavelengths, not longer wavelengths. So **D is incorrect**. --- 6. **Final conclusion** Correct options are: $$\boxed{A,\ B,\ C}$$ The stored correct answer is only $C$, which is incomplete/incorrect.

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View written solutionFree

Correct answer: A, B, C

Use Stefan–Boltzmann law for net radiative loss

A body at temperature $T$ in surroundings at temperature $T_0$ loses energy by radiation at the net rate

$P = \sigma A\left(T^4 - T_0^4\right).$

Here,

$A \approx 1\,\text{m}^2$
$T_0 = 300\,\text{K}$
body temperature $T = 310\,\text{K}$
given $\sigma T_0^4 = 460\,\text{W m}^{-2}$

So we evaluate each option.

Option A

We need the energy radiated in $1\,$ s, i.e. net power in watts.

$P = \sigma A(310^4 - 300^4).$

Since the temperature difference is small, use linearization:

$T^4 - T_0^4 \approx 4T_0^3(T-T_0).$

Thus,

$P \approx 4\sigma A T_0^3(10).$

Now from

$\sigma T_0^4 = 460,$

we get

$\sigma T_0^3 = \frac{460}{300}.$

Hence

= \frac{18400}{300} \approx 61.3\,\text{W}.$$ So in $1\,$s, energy radiated is about $$61\,\text{J},$$ which is close to $60\,$J. So **A is correct**. --- 3. **Option B** Initially, $$P = \sigma A(T^4 - T_0^4).$$ If surrounding temperature decreases by a small amount $\Delta T_0$, with body temperature kept same, then the new loss becomes $$P' = \sigma A\left(T^4 - (T_0-\Delta T_0)^4\right).$$ Increase in radiative loss: $$\Delta W = P' - P = \sigma A\left[T_0^4 - (T_0-\Delta T_0)^4\right].$$ For small $\Delta T_0$, $$(T_0-\Delta T_0)^4 \approx T_0^4 - 4T_0^3\Delta T_0.$$ Therefore, $$\Delta W \approx 4\sigma A T_0^3\Delta T_0.$$ Since $A = 1\,\text{m}^2$ here, $$\Delta W = 4\sigma T_0^3\Delta T_0.$$ So **B is correct**. --- 4. **Option C** Net radiative loss is $$P = \sigma A(T^4 - T_0^4).$$ At fixed body temperature $T$ and fixed surroundings $T_0$, the loss is directly proportional to exposed area $A$. So if a person curls up and reduces exposed surface area, then radiative energy loss decreases while body temperature can be maintained more easily. So **C is correct**. --- 5. **Option D** By Wien's displacement law, $$\lambda_{\max} T = \text{constant}.$$ If body temperature rises, then $T$ increases, so $$\lambda_{\max} \propto \frac{1}{T}$$ decreases. Thus the peak shifts to **shorter** wavelengths, not longer wavelengths. So **D is incorrect**. --- 6. **Final conclusion** Correct options are: $$\boxed{A,\ B,\ C}$$ The stored correct answer is only $C$, which is incomplete/incorrect.

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