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Q10.20: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.20 Solution: A body cools from 80 �C to 50 �C in 5 minutes. Calculate the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently replace the differential form $\frac{dT}{dt}=-k(T-T_s)$ with the algebraic ratio $\frac{\Delta T}{\Delta t}=k\,\overline{(T-T_s)}$, treating the cooling as linear over the interval. This ignores the exponential nature of Newton's law; the correct step is $\ln\frac{T-T_s}{T_0-T_s}=-kt$, then solve for $k$ and the required time.

A body cools from 80 �C to 50 �C in 5 minutes. Calculate the time it takes to cool from 60 �C to 30 �C. The temperature of the surroundings is 20 �C.

Step-by-Step Verified Solution

**Answer:** t=9t = 9 min **Step-by-Step Explanation:** Using Newton's law of cooling:
dTdt=−k(T−Ts)\frac{dT}{dt} = -k(T - T_s)
For the first case: T1=80∘CT_1 = 80^\circ\text{C}, T2=50∘CT_2 = 50^\circ\text{C}, t1=5t_1 = 5 min, Ts=20∘CT_s = 20^\circ\text{C}. Average temperature = 80+502=65∘C\frac{80 + 50}{2} = 65^\circ\text{C}. Temperature difference = 65−20=45∘C65 - 20 = 45^\circ\text{C}.
80−505=k×45  ⟹  6=45k  ⟹  k=645=0.1333 min−1\frac{80 - 50}{5} = k \times 45 \implies 6 = 45k \implies k = \frac{6}{45} = 0.1333 \text{ min}^{-1}
For the second case: T1=60∘CT_1 = 60^\circ\text{C}, T2=30∘CT_2 = 30^\circ\text{C}, t2=?t_2 = ? Average temperature = 60+302=45∘C\frac{60 + 30}{2} = 45^\circ\text{C}. Temperature difference = 45−20=25∘C45 - 20 = 25^\circ\text{C}.
60−30t2=0.1333×25  ⟹  30t2=3.333  ⟹  t2=303.333=9 min\frac{60 - 30}{t_2} = 0.1333 \times 25 \implies \frac{30}{t_2} = 3.333 \implies t_2 = \frac{30}{3.333} = 9 \text{ min}

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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