A solid body X of heat capacity C is kept in an atmosphere whose temperature is T_A = 300 K. At time

Q: Why do we use Newton's law of cooling for radiation here?

Newton's law of cooling is an approximation for heat loss due to radiation (and natural convection) when the temperature difference between the body and the surroundings is relatively small.

Q: Why is the temperature of body Y considered constant?

The problem states that the heat capacity of body Y is 'so large that any variation in its temperature may be neglected.' This makes it act as an infinite thermal reservoir.

Q: How do we combine the cooling rates in the second phase?

Since the heat loss mechanisms (radiation and conduction) are independent and occur simultaneously, their rates of heat loss simply add up. The total rate of change of temperature is proportional to the total rate of heat loss.

Q: Why is the time interval for the second integration $2t_1$?

The second phase starts at $t = t_1$ and we need to find the temperature at $t = 3t_1$. The duration of this phase is $3t_1 - t_1 = 2t_1$.

Q: What happens to the $\ln(2)$ term in the final step?

We substitute $kt_1 = \ln(2)$ into the term $-2kt_1$, which becomes $-2\ln(2)$. Using logarithm properties, $-2\ln(2) = \ln(2^{-2}) = \ln(1/4)$. When we exponentiate, this $\ln(1/4)$ becomes a factor of $1/4$, which multiplies with the $50$ to give $12.5$.

Question

A solid body $X$ of heat capacity $C$ is kept in an atmosphere whose temperature is $T_A = 300	ext{ K}$. At time $t = 0$, the temperature of $X$ is $T_0 = 400	ext{ K}$. It cools according to Newton's law of cooling. At time $t_1$ its temperature is found to be $350	ext{ K}$.
At this time ($t_1$) the body $X$ is connected to a large body $Y$ at atmospheric temperature $T_A$ through a conducting rod of length $L$, cross-sectional area $A$ and thermal conductivity $K$. The heat capacity of $Y$ is so large that any variation in its temperature may be neglected. The cross-sectional area $A$ of the connecting rod is small compared to the surface area of $X$. Find the temperature of $X$ at time $t = 3t_1$.

Sigma Percentile · Accepted Answer

The problem of a cooling body might seem straightforward at first glance, but this particular challenge from the JEE Advanced archives introduces a beautiful twist. We are not just dealing with a single mode of heat transfer; we are witnessing a transition from pure radiation to a parallel combination of radiation and conduction. Let's embark on this thermal journey and uncover the mathematics behind it!

### Phase 1: The Solitary Cooling

Imagine our solid body $X$ sitting alone in the atmosphere. From time $t = 0$ to $t = t_1$, it is cooling down purely by radiating heat to its surroundings. The atmosphere acts as a massive thermal reservoir, maintaining a constant temperature of $T_A = 300	ext{ K}$.

According to Newton's law of cooling, the rate at which the body's temperature drops is directly proportional to the temperature difference between the body and the atmosphere. We can express this mathematically as:

$$-\frac{dT}{dt} = k(T - T_A)$$

Here, $k$ is the cooling constant that depends on the body's properties and surface area.

### The Mathematics of Newton's Law

To find out exactly how much the body cools over time, we need to integrate this differential equation. By separating the variables, we bring the temperature terms to one side and the time terms to the other:

$$\frac{dT}{T - T_A} = -k dt$$

Now, we integrate from the initial state at $t = 0$ (where $T_0 = 400	ext{ K}$) to the state at $t = t_1$ (where $T_1 = 350	ext{ K}$):

$$\int_{T_0}^{T_1} \frac{dT}{T - T_A} = -k \int_0^{t_1} dt$$

Evaluating the integrals gives us a logarithmic relation:

$$\ln\left(\frac{T_1 - T_A}{T_0 - T_A}ight) = -k t_1$$

Let's plug in the known temperature values to evaluate the product $kt_1$:

$$-k t_1 = \ln\left(\frac{350 - 300}{400 - 300}ight) = \ln\left(\frac{50}{100}ight) = \ln\left(\frac{1}{2}ight) = -\ln(2)$$

This beautifully simplifies to a crucial piece of information that we will need later:

$$k t_1 = \ln(2)$$

### Phase 2: The Parallel Pathways

Now begins the second phase! At time $t_1$, the scenario changes. We connect body $X$ to a massive body $Y$ using a conducting rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$.

Because body $Y$ is so massive, its temperature remains constant at $T_A = 300	ext{ K}$, just like the atmosphere. This means body $X$ is now losing heat through two parallel paths simultaneously:
1. **Radiation** to the surrounding air.
2. **Conduction** through the rod to body $Y$.

Let's figure out the cooling rate just due to this new conduction path. The rate of heat flow through the rod is given by Fourier's law of heat conduction:

$$\frac{dQ}{dt} = \frac{KA(T - T_A)}{L}$$

Since the heat lost by the body is related to its temperature drop by $dQ = -C dT$ (where $C$ is the heat capacity), we can express the cooling rate due to conduction as:

$$\left(-\frac{dT}{dt}ight)_{	ext{cond}} = \frac{KA}{LC}(T - T_A)$$

Because these two cooling mechanisms are happening at the same time, the total, net rate of cooling is simply the sum of the individual cooling rates:

$$\left(-\frac{dT}{dt}ight)_{	ext{net}} = k(T - T_A) + \frac{KA}{LC}(T - T_A)$$

Factoring out the temperature difference, we get our new, effective differential equation for the second phase:

$$-\frac{dT}{dt} = \left(k + \frac{KA}{LC}ight)(T - T_A)$$

### The Final Integration

With our new differential equation ready, we integrate it again. This time, we are integrating from time $t_1$ to $3t_1$. The duration of this phase is $\Delta t = 3t_1 - t_1 = 2t_1$. The temperature goes from $T_1$ to our unknown final temperature $T_2$.

$$\int_{T_1}^{T_2} \frac{dT}{T - T_A} = -\left(k + \frac{KA}{LC}ight) \int_{t_1}^{3t_1} dt$$

$$\ln\left(\frac{T_2 - T_A}{T_1 - T_A}ight) = -\left(k + \frac{KA}{LC}ight)(2t_1)$$

Finally, let's substitute all our knowns into this equation. We plug in $T_1 = 350	ext{ K}$, $T_A = 300	ext{ K}$, and remember that $kt_1 = \ln(2)$:

$$\ln\left(\frac{T_2 - 300}{350 - 300}ight) = -2kt_1 - \frac{2KAt_1}{LC}$$

$$\ln\left(\frac{T_2 - 300}{50}ight) = -2\ln(2) - \frac{2KAt_1}{LC}$$

Using the properties of logarithms, we know that $-2\ln(2) = \ln(2^{-2}) = \ln(1/4)$. So the equation becomes:

$$\ln\left(\frac{T_2 - 300}{50}ight) = \ln\left(\frac{1}{4}ight) - \frac{2KAt_1}{LC}$$

To isolate $T_2$, we exponentiate both sides:

$$\frac{T_2 - 300}{50} = \frac{1}{4} e^{-\frac{2KAt_1}{LC}}$$

Multiplying by 50 and adding 300, we arrive at our final, elegant result:

**$$T_2 = 300 + 12.5 e^{-\frac{2KAt_1}{LC}}$$**

And there we have it! The final temperature of body $X$ at time $3t_1$, perfectly capturing the combined effects of radiation and conduction.

Visualized Solution

$N e w t o n^{'} s L a w o f C oo l in g$

$I n t e g r a t in g t h e C oo l in g E q u a t i o n$

$E v a l u a t in g t h e C oo l in g C o n s t an t$

$C oo l in g w i t h C o n d u c t i o n$

$R a t eo f H e a t L oss v ia C o n d u c t i o n$

$N e tR a t eo f C oo l in g$

$I n t e g r a t in g t h e N e tC oo l in g E q u a t i o n$

$S o l v in g f or F ina l T e m p er a t u r e$

The Sigma Insight: Heat Transfer

Phase 1

The Mathematics of Newton's Law

Phase 2

The Final Integration

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

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

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

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

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

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