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Thermal Properties of Matter - Newton's Law of Cooling

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Newton's Law of Cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small and the nature of the radiating surface remains the same.

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Mathematically, if TT is the temperature of the body and TsT_s is the temperature of the surroundings, then the rate of cooling is given by dTdt=−K(T−Ts)\frac{dT}{dt} = -K(T - T_s), where KK is a positive constant depending on the surface area and nature of the surface.

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For practical calculations involving a body cooling from temperature T1T_1 to T2T_2 in time tt, we use the average temperature form: T1−T2t=K(T1+T22−Ts)\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_s\right).

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The law is an approximation of Stefan's Law and is strictly valid only for small temperature differences (typically less than 35∘C35^{\circ}C to 40∘C40^{\circ}C).

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The graph of temperature TT versus time tt is an exponential decay curve, while a graph of log⁡(T−Ts)\log(T - T_s) versus time tt is a straight line with a negative slope.

📐Formulae

dQdt=−k(T−Ts)\frac{dQ}{dt} = -k(T - T_s) (Rate of heat loss)

dTdt=−K(T−Ts)\frac{dT}{dt} = -K(T - T_s) (Rate of fall of temperature)

T1−T2t=K[T1+T22−Ts]\frac{T_1 - T_2}{t} = K\left[\frac{T_1 + T_2}{2} - T_s\right] (Average form for numericals)

ln⁡(T−Ts)=−Kt+C\ln(T - T_s) = -Kt + C (Logarithmic form)

T=Ts+(T0−Ts)e−KtT = T_s + (T_0 - T_s)e^{-Kt} (Temperature at any time tt)

💡Examples

Problem 1:

A body cools from 60∘C60^{\circ}C to 50∘C50^{\circ}C in 1010 minutes. If the room temperature is 25∘C25^{\circ}C, find the time it will take to cool from 50∘C50^{\circ}C to 40∘C40^{\circ}C.

Solution:

Using the formula T1−T2t=K(T1+T22−Ts)\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_s\right):

Case 1: T1=60∘C,T2=50∘C,t=10 min,Ts=25∘CT_1 = 60^{\circ}C, T_2 = 50^{\circ}C, t = 10 \text{ min}, T_s = 25^{\circ}C 60−5010=K(60+502−25)\frac{60 - 50}{10} = K\left(\frac{60 + 50}{2} - 25\right) 1=K(55−25)  ⟹  1=30K  ⟹  K=130 min−11 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30} \text{ min}^{-1}

Case 2: T1=50∘C,T2=40∘C,t=?,Ts=25∘CT_1 = 50^{\circ}C, T_2 = 40^{\circ}C, t = ?, T_s = 25^{\circ}C 50−40t=K(50+402−25)\frac{50 - 40}{t} = K\left(\frac{50 + 40}{2} - 25\right) 10t=130(45−25)\frac{10}{t} = \frac{1}{30}(45 - 25) 10t=2030  ⟹  10t=23\frac{10}{t} = \frac{20}{30} \implies \frac{10}{t} = \frac{2}{3} 2t=30  ⟹  t=15 minutes2t = 30 \implies t = 15 \text{ minutes}

Explanation:

We first calculate the cooling constant KK using the data from the first interval. Then, we apply this constant to the second interval to find the unknown time. Note that it takes longer (1515 min vs 1010 min) to cool through the same temperature range as the body approaches the surrounding temperature.

Newton's Law of Cooling - Revision Notes & Key Formulas | CBSE Class 11 Physics