A human body has a surface area of approximately 1 m^2. The normal body temperature is 10K above the

Q: Why did we calculate the *net* energy radiated instead of just the emitted energy for option C?

This is a classic JEE ambiguity! The phrase "energy radiated" technically means the total energy emitted ($\sigma A T^4$). However, the context of maintaining body temperature implies we care about the *heat loss*, which is the net radiation. The answer key confirms they intended for you to calculate the net energy.

Q: Why can we use the binomial approximation here?

The body temperature is $310\text{ K}$ and the room is $300\text{ K}$. The ratio $10/300$ is $1/30$, which is much less than 1. Whenever $x \ll 1$, $(1+x)^n \approx 1+nx$ is a highly accurate and time-saving tool!

Q: In option B, why is $dT_0$ taken as negative?

The option states that the surrounding temperature *reduces*. A reduction means the change is negative, so $dT_0 = -\Delta T_0$. This cancels the negative sign from the derivative, resulting in a positive increase in required power.

Q: Does curling up really reduce heat loss?

Absolutely! The net power radiated is directly proportional to the exposed surface area. By curling up, you hide parts of your body (like your chest and stomach) from the cold surroundings, effectively reducing $A$ and minimizing heat loss.

Q: What if the temperature difference was large? Could we still use the binomial approximation?

No. If the temperature difference was, say, $100\text{ K}$, the ratio would be $100/300 = 1/3$. The binomial approximation would introduce a significant error. In such cases, you must calculate the exact fourth powers.

Question

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

Options:

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

(B) If the surrounding temperature reduces by a small amount $\Delta T_0 \ll T_0$, 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

(C) The amount of energy radiated by the body in 1s is close to 60 J

(D) Reducing 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

Sigma Percentile · Accepted Answer

### The Physics of Staying Warm: Decoding Human Body Radiation

Have you ever wondered why you instinctively curl up into a ball when you feel cold? Or how much energy your body is constantly radiating into the room around you? This fascinating JEE Advanced problem takes us on a journey through the physics of thermal radiation, blending Wien's Displacement Law, the Stefan-Boltzmann Law, and a clever mathematical approximation.

Let's break down the options one by one and uncover the physical realities behind the equations.

### Analyzing Option A: Wien's Displacement Law

Option (a) suggests that if the body temperature rises, the peak of the emitted electromagnetic spectrum shifts to longer wavelengths. To verify this, we must recall **Wien's Displacement Law**, which states that the product of the peak wavelength $\lambda_m$ and the absolute temperature $T$ is a constant:

$$\lambda_m T = b$$

This inverse relationship means that as the temperature $T$ increases, the peak wavelength $\lambda_m$ must *decrease*. The radiation shifts towards shorter wavelengths (higher frequencies and higher energies). Therefore, option (a) is physically incorrect.

### Analyzing Option C: The Ambiguity of "Energy Radiated"

Option (c) asks us to calculate the energy radiated by the body in 1 second. Here lies a classic JEE ambiguity! Technically, the total energy emitted by the body is $P_{	ext{emit}} = \sigma A T^4$. However, the body is also absorbing radiation from the room at $T_0$. In the context of maintaining body temperature, we are interested in the **net heat loss**.

The net power radiated is given by the Stefan-Boltzmann Law:

$$P_{	ext{net}} = \sigma A (T^4 - T_0^4)$$

We are given that the body temperature is $10	ext{ K}$ above the room temperature, so $T = T_0 + 10$. Substituting this into our equation:

$$P_{	ext{net}} = \sigma A ((T_0 + 10)^4 - T_0^4)$$

To make this calculation elegant, we factor out $T_0^4$:

$$P_{	ext{net}} = \sigma A T_0^4 \left[ \left(1 + \frac{10}{T_0}ight)^4 - 1 ight]$$

Since $10 \ll 300$, the term $\frac{10}{T_0}$ is very small. This is the perfect scenario to deploy the **Binomial Approximation**: $(1+x)^n \approx 1+nx$.

$$P_{	ext{net}} \approx \sigma A T_0^4 \left[ 1 + 4\left(\frac{10}{T_0}ight) - 1 ight] = \sigma A T_0^4 \left( \frac{40}{T_0} ight)$$

Now, we simply plug in the given values: $A = 1 	ext{ m}^2$, $\sigma T_0^4 = 460 	ext{ Wm}^{-2}$, and $T_0 = 300 	ext{ K}$:

$$P_{	ext{net}} \approx 460 	imes 1 	imes \left( \frac{40}{300} ight) = 460 	imes \frac{4}{30} \approx 61.33 	ext{ W}$$

Since power is energy per second, the net energy radiated in 1 second is approximately $60 	ext{ J}$. Option (c) is correct!

### Analyzing Option B: Differentiating the Power Equation

Option (b) asks how much extra energy the body must radiate if the room temperature $T_0$ drops by a small amount $\Delta T_0$, assuming the body temperature $T$ remains constant.

To find the change in power, we differentiate our net power equation with respect to $T_0$:

$$dP_{	ext{net}} = \sigma A (0 - 4T_0^3 dT_0)$$

Since the room temperature is *reducing*, the change $dT_0$ is negative, meaning $dT_0 = -\Delta T_0$. Substituting this in:

$$\Delta W = dP_{	ext{net}} = -4\sigma A T_0^3 (-\Delta T_0) = 4\sigma A T_0^3 \Delta T_0$$

Given that the area $A = 1 	ext{ m}^2$, the extra energy required per unit time is exactly $4\sigma T_0^3 \Delta T_0$. Option (b) is flawlessly correct.

### Analyzing Option D: The Physics of Curling Up

Finally, option (d) states that reducing the exposed surface area helps maintain body temperature. Looking back at our master equation:

$$P_{	ext{net}} = \sigma A (T^4 - T_0^4)$$

It is clear that the net power lost is directly proportional to the exposed surface area $A$. When you curl up, you hide parts of your body from the cold surroundings, effectively reducing $A$. This minimizes the heat lost via radiation, helping your body conserve energy and stay warm. Option (d) is a beautiful real-world application of the math and is absolutely correct.

**Final Conclusion:** The correct options are (b), (c), and (d).

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Visualized Solution

$W i e n^{'} s D i s pl a ce m e n t L a w$

$N e t P o w er R a d ia t e d$

$S u b s t i t u t i o n an d F a c t or in g$

$B in o mia l A pp r o x ima t i o n$

$F ina l C a l c u l a t i o n f or O pt i o n C$

$D i f f er e n t ia t in g P o w er E q u a t i o n$

$E x t r a P o w er R e q u i r e d$

$E f f ec t o f S u r f a ce A r e a$

The Sigma Insight: Heat Transfer

The Physics of Staying Warm

Analyzing Option A

Analyzing Option C

Analyzing Option B

Analyzing Option D

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wien′s Displacement Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Power Radiated

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution and Factoring

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binomial Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation for Option C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating Power Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extra Power Required

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effect of Surface Area