Newton’s Law of Cooling in 59 Sec– Concept, Formula & JEE Tips

 

📌 Newton’s Law of Cooling – Explained for JEE | Formula + Graph + Example

---

🖼️ Image:

🔢 Formula:

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

Where:

• $T$ = Temperature of the object

• $Tₐ$ = Ambient temperature (surroundings)

• $k$ = Cooling constant (depends on material and conditions)

• $\frac{dT}{dt}$ = Rate of cooling

---

🔍 Concept Points:

• Cooling rate ∝ Temperature difference between object and environment

• Greater the difference, faster the cooling

• As object cools down, rate decreases → slows down as it nears $T_{\mathrm{a}}$

---

🔢 Example:

Case 1:

• $T = 90^\circ C$, $Tₐ = 30^\circ C$
  → High difference = fast cooling

Case 2:

• $T = 55^\circ C$, $Tₐ = 50^\circ C$
  → Small difference = slow cooling

---

📊 Full Equation (JEE Form):

$$T = Tₐ + (T₀ - Tₐ) \cdot e^{-kt}$$

Where:

• $T₀$ = Initial temperature

• $T$ = Temperature at time $t$
  

• $e$ = Exponential base

⚠️ The minus sign indicates cooling (temperature decreases over time).

---

🎥 Video Solution:

---

🧠 JEE Tip:

• Expect graph-based questions on T vs t

• Learn the ratio method for solving problems without full derivation

• Sometimes they ask about final temperature, time required, or rate at a given instant

---

🧠 Pro Tip:

Break down long expressions during revision. Newton's Law often appears in Thermodynamics or Modern Physics sections — practice derivations!

---

📩 Have a Doubt?

Drop your doubt at 📧 doubtifyqueries@gmail.com or DM on Instagram @DoubtifyJEE