Cooling Law Formula and Applications for JEE

❓ Concept Question

What is Newton’s Law of Cooling and how does cooling rate depend on surrounding temperature?

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🖼 Concept Image

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✍️ Short Concept

This concept is based on:

👉 Cooling of hot bodies
👉 Temperature difference concept
👉 Exponential cooling relation.

Main idea:

$$\text{Cooling rate} \propto (T-T_s)$$

where:

• $T$ = temperature of body
• $T_s$ = surrounding temperature

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🔷 Step 1 — Newton’s Law of Cooling 💯

Cooling equation:

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

Where:

• $k$ = cooling constant
• $\frac{dT}{dt}$= rate of temperature change

Negative sign indicates:

$$\text{Temperature decreases with time}$$

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🔷 Step 2 — Main Cooling Idea

According to Newton’s Law:

$$\text{Cooling rate} \propto (T-T_s)$$

So:

✔ Larger temperature difference → faster cooling

✔ Smaller temperature difference → slower cooling

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🔷 Step 3 — Example Understanding

Case 1:

$$T=90^\circ C,\quad T_s=30^\circ C$$

Temperature difference:

$$60^\circ C$$

Hence cooling is fast.

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Case 2:

$$T=55^\circ C,\quad T_s=50^\circ C$$

Temperature difference:

$$5^\circ C$$

Hence cooling is slow.

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🔷 Step 4 — Full Cooling Formula

Integrated form:

$$T=T_s+(T_0-T_s)e^{-kt}$$

Where:

• $T_0$ = initial temperature
• $T$ = temperature after time $t$
  

This shows exponential cooling.

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Minus sign ignore kar dena

❌ Cooling rate ko directly temperature proportional maan lena

❌ Surrounding temperature effect bhool jaana

Remember:

$$\boxed{ \text{Cooling depends on }(T-T_s) }$$

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✅ Final Takeaway

For Newton’s Law of Cooling:

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

Greater the temperature difference:

$$\Rightarrow \text{Faster cooling}$$

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⭐ Golden JEE Insight

As body temperature approaches surrounding temperature:

$$T\to T_s$$

then:

$$\frac{dT}{dt}\to0$$

So cooling gradually slows down.

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📚 Related Topics

• Silver Halide Formation in Organic Analysis
• Strong Field vs Weak Field Ligands Explained
• Faraday Law and Nernst Equation Combined Concept