Newton's Law Of Cooling

Newton's Law of Cooling describes how the Temperature of an object changes when it is in contact with a different surrounding temperature.  Law is helpful in understanding heat transfer phenomena. 

1.0Statement of Newton's Law of Cooling

Newton's Law of Cooling asserts that the speed at which an object's temperature changes is directly linked to the disparity between its Temperature. The surrounding ambient Temperature relationship dictates how quickly an object cools or heats as it interacts thermally with its environment.

Critical points of Newton's Law of Cooling:

• The rate at which the temperature changes varies directly with the temperature difference between the object and its surrounding environment.
• Objects with a more significant temperature difference are incredible (or heat, depending on context) faster.
• As the object approaches thermal equilibrium with its surroundings, the rate of temperature change decreases.

2.0Examples of Newton's Law of Cooling

(1). Hot Beverages: When a cup of hot coffee is left on a table in a room with cooler ambient Temperature, It Gradually cools down. The rate at which it cools follows Newton's Law of Cooling, with the temperature difference between the coffee and the room determining how quickly it reaches room temperature.

(2).Cooking: When hot food is taken out of an oven and left to cool on a countertop, its Temperature decreases over time following Newton's Law of Cooling. The rate at which the food cools is influenced by the temperature difference between the food and the room.

(3).Heating and Air Conditioning: According to Newton's Law of Cooling, homes and buildings cool down or warm up. For instance, a house loses heat to the outside air during winter. 

(4).Electronics: Electronic devices, such as laptops or smartphones, heat up during use. Once turned off, they begin to cool down following Newton's Law of Cooling. The temperature difference between the device and the room influences the cooling rate.

3.0Limitations Newton's Law of Cooling

(1) Heat loss from an object occurs solely through radiation.

(2) The change in temperature between the object and its surroundings should be minimal. 

(3) The ambient Temperature remains constant throughout the cooling process.

4.0Theory of Newton’s  Law of Cooling

Newton's Law of Cooling provides a mathematical model for how the temperature of an object changes over time when it is exposed to a different ambient temperature. The law states that the rate at which an object's temperature changes is proportional to the difference between its temperature and the surrounding ambient temperature.

Key aspects of Newton's Law of Cooling include:

• Proportional Relationship: The rate at which an object cools or heats is directly proportional to the temperature difference between the object itself and the surrounding environment.
• Temperature Difference Impact: Objects with more considerable temperature differences cool (or heat) more quickly.
• Approaching Equilibrium: As the object's Temperature approaches the ambient temperature, the rate of Temperature change decreases.

5.0Derivation of Newton's Law of Cooling

According to this law, the rate of loss of heat, $\left(-\frac{dQ}{dt}\right)$ of the body is directly proportional to the change in  temperature ∆T = (T – T₀ ) of the body and the surroundings.The law holds good only for slight differences of temperature.

$-\frac{dQ}{dt}\propto \left(\mathrm{T}-{\mathrm{T}}_0\right)$

$-\frac{dQ}{dt}=\mathrm{K}\left(\mathrm{T}-{\mathrm{T}}_0\right)$. (∴ dQ = msdT ; dT = Fall in Temperature)

$-\frac{msdT}{dt}=\mathrm{K}\left(\mathrm{T}-{\mathrm{T}}_0\right)$

$\frac{dT}{T-T_0}=-\mathrm{Kdt}$  (∴K  is a constant)

Integrate both the sides

$\int \frac{dT}{T-T}=\int -\mathrm{Kdt}$

log_e(T - T₀) = -Kt + C

$\mathrm{T}-{\mathrm{T}}_0=e^{-kt+c}$

T = T₀ + e^-Kt + C. ………...This equation calculates the time of cooling of a body through a particular range of temperature.

Note-For Numerical Problems (Newton’s Law of Cooling)   

$\frac{T_2-T_1}{t}=-\mathrm{K}\left[\frac{T_1+T_2}{2}-T_0\right]$

Graph between temperature(T) and time(t):

Graph between log (T - T₀)  versus time(t):

6.0Solved Questions on Newton's Law of Cooling  

Q-1. A cup of tea cools from 80⁰ C to  60⁰ C  in  one minute. The ambient Temperature is  30⁰ C. In cooling from 60⁰ C to 50⁰ C it will take time?

Sol.

$\frac{T_1-T_2}{t}\propto \left(\frac{T_1+T_2}{2}-T\right)$

Case 1. $\frac{80-60}{60}\propto \left(\frac{80+60}{2}-30\right)$ …………(1)

Case 2. $\frac{60-50}{t}\propto \left(\frac{60+50}{2}-30\right)$ ………….(2)

From equation (1) and (2) it will take 48 seconds.

Q-2. It takes T minutes for a body to cool down from 62 ⁰C to 61⁰C when the surrounding temperature is 30 ⁰C. The time taken by the body to cool from 46⁰C to 45.5⁰c is.

Sol.  

$\frac{T_1-T_2}{t}\propto \left(\frac{T_1+T_2}{2}-T\right)$

Case 1. $\frac{62-61}{T}\propto \left(\frac{62+61}{2}-30\right)$

Case  2. $\frac{46-45.5}{t}\propto \left(\frac{46+45.5}{2}-30\right)$

From case 1 and case 2 time is equal to T seconds.

Q-3. A beaker  full of hot water cools from 75⁰C TO 70⁰C in time T₁, from 70⁰C to 65⁰C in time T₂ and from 65⁰C to 60⁰C in time T₃. Arrange the following which cools faster?

Sol. The rate at which a body cools decreases as the temperature difference between the body and its surroundings decreases. Consequently, the time required for the body to cool increases. Stated differently, a hotter body cools down more rapidly T₁>T₂>T₃ .