The temperature T(t) of a body at time t = 0 is 160° F and it decreases continuously as per the diff

Q: Why can't I just use linear interpolation? If it drops 40 degrees in 15 minutes, shouldn't it drop another 40 in the next 15?

That is a very common trap! Linear interpolation assumes a constant rate of change. But Newton's Law of Cooling tells us the rate of change is proportional to the temperature difference. As the body cools, the difference decreases, so the rate of cooling slows down. It is an exponential decay, not a linear one. The body cools rapidly at first and then 'lingers' as it approaches the room temperature.

Q: Do I really need to find the value of $K$?

Absolutely not! In JEE Advanced, time is your most precious resource. By recognizing that $45 = 3 \times 15$, we can use the properties of exponents to solve the problem using ratios. Calculating $K$ involves logarithms and potential rounding errors—avoid it whenever you can. Work with the decay factor $e^{-Kt}$ directly.

Q: What if the temperature $T$ was less than 80? Would the math still hold?

Yes, the math is robust. If $T < 80$, then $(T - 80)$ is negative, and the rate of change $\frac{dT}{dt}$ becomes positive, meaning the body would warm up towards the ambient temperature. The differential equation naturally handles both cooling and warming scenarios!

Question

The temperature $T(t)$ of a body at time $t = 0$ is $160^\circ F$ and it decreases continuously as per the differential equation $\frac{dT}{dt} = -K(T - 80)$, where $K$ is positive constant. If $T(15) = 120^\circ F$, then $T(45)$ is equal to

Options:

(A) $85^\circ F$

(B) $95^\circ F$

(C) $90^\circ F$

(D) $80^\circ F$

Sigma Percentile · Accepted Answer

### The Symphony of Cooling: Understanding Newton's Law

Imagine you have just brewed a steaming cup of tea. You place it on your desk, and as you sit down to study, you notice it cooling.

It does not drop in temperature at a constant rate. It cools rapidly at first, and then, as it gets closer to the temperature of the room, the cooling slows down, almost as if the tea is 'hesitating' to reach room temperature.

This is the physical reality described by **Newton's Law of Cooling**, and it is a beautiful example of exponential decay.

### The Differential Equation: The Language of Change

In our problem, we are given the differential equation:

$$\frac{dT}{dt} = -K(T - 80)$$

The term $\frac{dT}{dt}$ represents the rate of change of temperature with respect to time. The right side, $-K(T - 80)$, tells us that the rate of cooling is directly proportional to the difference between the current temperature $T$ and the ambient temperature, which is $80^{\circ}F$.

This $80^{\circ}F$ is our asymptote. Mathematically, as $t 	o \infty$, $T 	o 80$. The body will never actually reach $80^{\circ}F$ in finite time, but it will get infinitely close.

### The Calculus of Cooling

To solve this, we use the method of separation of variables. We group all the $T$ terms on one side and all the $t$ terms on the other:

$$\frac{dT}{T - 80} = -K dt$$

Now, we integrate both sides:

$$\int \frac{dT}{T - 80} = \int -K dt$$

The left side integrates to the natural logarithm, $\ln(T - 80)$, and the right side gives us $-Kt + C$.

Exponentiating both sides, we arrive at the general solution:

$$T(t) - 80 = e^{-Kt + C} = e^C \cdot e^{-Kt}$$

Let $A = e^C$. Then $T(t) - 80 = A e^{-Kt}$. We find $A$ using our initial condition at $t = 0$, where $T = 160^{\circ}F$:

$$160 - 80 = A e^0 \Rightarrow A = 80$$

Thus, our specific temperature function is:

$$T(t) - 80 = 80 e^{-Kt}$$

### The JEE Mindset: Efficiency Over Brute Force

Now, we reach the intermediate condition: at $t = 15$, $T = 120^{\circ}F$. Most students would immediately try to solve for $K$.

They would write $120 - 80 = 80 e^{-15K}$, which simplifies to $40 = 80 e^{-15K}$, or $e^{-15K} = \frac{1}{2}$. They might then take the natural log to find $K = \frac{\ln(2)}{15}$.

**Stop!** Do not do this. In the heat of a JEE Advanced exam, calculating $K$ is a waste of time and an invitation to arithmetic errors.

Look at the target: we need to find $T$ at $t = 45$. Notice the relationship between the times: $45 = 3 	imes 15$. This is not a coincidence; it is a design feature of the problem.

### The Power of Exponents

We want to find $T(45)$. Our equation is $T(45) - 80 = 80 e^{-45K}$.

Using the laws of exponents, we can rewrite $e^{-45K}$ as $(e^{-15K})^3$. We already know that $e^{-15K} = \frac{1}{2}$.

Therefore:

$$e^{-45K} = \left(\frac{1}{2}ight)^3 = \frac{1}{8}$$

This is the elegance of the method. We have bypassed the need to calculate $K$ entirely. We simply substitute this value back into our equation:

$$T(45) - 80 = 80 	imes \frac{1}{8}$$

### The Final Calculation

Calculating the right side is trivial: $80 	imes \frac{1}{8} = 10$. So, $T(45) - 80 = 10$.

Adding $80$ to both sides, we get:

**$T(45) = 90^{\circ}F$**

Look at what we have achieved. We started with a differential equation, navigated through the calculus of integration, and finished with a clean, logical deduction using the properties of exponents.

Remember, in physics and mathematics, the most elegant path is usually the one that respects the structure of the equation. Keep this mindset, and you will find that even the most complex problems become manageable.

Select Answer:

Visualized Solution

Newton's Law of Cooling

Variable Separation

Integration

Initial Condition at $t = 0$

General Solution for $T (t)$

Intermediate Condition at $t = 15$

Finding the Decay Factor

Target Condition at $t = 45$

The Power Trick

Substitution

Final Calculation

The Sigma Insight: Variable Separable Method

The Symphony of Cooling

The Differential Equation

The Calculus of Cooling

The JEE Mindset

The Power of Exponents

The Final Calculation

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