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Newton’s Law of Cooling

Newton’s Law of Cooling :- The rate at which a body loses heat is directly proportional to the difference between the temperature of the body and the temperature of its surroundings, provided that this temperature difference is small (typically less than ).

$\displaystyle -\frac{dQ}{dt}\propto (T-{{T}_{0}})$

$\displaystyle \Rightarrow -\frac{dQ}{dt}=k(T-{{T}_{0}})$ …..(1)

When $\displaystyle \text{ }\!\![\!\!\text{ }(T-{{T}_{0}})\le 35{}^\circ C\text{ }\!\!]\!\!\text{ }$ .

Here -ve sign indicates loss of heat.

T = temperature of body [in °C]

T₀ = temperature of surrounding

(T – T₀) = excess temperature of the body over the surroundings

Also,

$\displaystyle \frac{dQ}{dt}=ms\frac{dT}{dt}$

$\displaystyle \Rightarrow -\frac{dT}{dt}=\frac{1}{ms}\frac{dQ}{dt}$ …..(2)

The negative sign means the temperature is falling as time passes.

Using equation (1) in equation (2), we get

$\displaystyle \frac{dT}{dt}=\frac{k}{ms}(T-{{T}_{0}})$

$\displaystyle \Rightarrow \frac{dT}{dt}=K(T-{{T}_{0}})$ …..(3)

Where $\displaystyle K=\frac{k}{ms}$ .

Excess temperature of the body over the surroundings

If the temperature of body decreases from T₁ to T₂ and temperature of surroundings is T₀ then,

Average temperature of the body = $\displaystyle \frac{{{T}_{1}}+{{T}_{2}}}{2}$

Average excess of temperature = $\displaystyle \left[ \frac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{0}} \right]$ …..(4)

Magnitude of rate of fall of temperature of the body = $\displaystyle \frac{dT}{dt}=\left[ \frac{{{T}_{1}}-{{T}_{2}}}{t} \right]$ …..(5)

Using equations (4) and (5) in equation (3), we get

$\displaystyle \left[ \frac{{{T}_{1}}-{{T}_{2}}}{t} \right]=K\left[ \frac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{0}} \right]$ …..(6)

Limitations of Newton’s Law of Cooling

The excess temperature of the object relative to the surroundings should not exceed 35 °C $\displaystyle \text{ }\!\![\!\!\text{ }(T-{{T}_{0}})\le 35{}^\circ C\text{ }\!\!]\!\!\text{ }$
Heat loss by the object should occur only by radiation.
The temperature of surroundings must remain constant during the cooling of the body.
Newton’s law of cooling is an extension of the Stefan–Boltzmann law.

Newton’s Law of Heating

If the temperature of the surroundings is hotter than the body’s temperature, the body will absorb heat from the surroundings, causing its temperature to rise. The form of the equation remains similar, but we have to swap the order of temperatures to reflect the increase in temperature :

$\displaystyle \left[ \frac{{{T}_{2}}-{{T}_{1}}}{t} \right]=K\left[ {{T}_{0}}-\frac{{{T}_{1}}+{{T}_{2}}}{2} \right]$ …..(7)

Derivation of Newton’s Law of Cooling from Steafan’s Boltzman Law

From Steafan-Boltzmann law, the rate of fall of temperature,

$\displaystyle {{R}_{F}}=\frac{d\theta }{dt}=\frac{\sigma A}{ms}({{T}^{4}}-T_{0}^{4})$

The temperature difference between the object and the surroundings,

$\displaystyle \Delta T=T-{{T}_{0}}$

$\displaystyle \Rightarrow T=\ {{T}_{0}}\ +\ \Delta T$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\frac{\sigma A}{ms}[{{\left( {{T}_{0}}\ +\ \Delta T \right)}^{\text{4}}}-T_{0}^{4}]$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\frac{\sigma A}{ms}\left[ T_{0}^{4}{{\left( 1\ +\ \frac{\Delta T}{{{T}_{0}}} \right)}^{\text{4}}}-T_{0}^{4} \right]$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\frac{\sigma A}{ms}T_{0}^{4}\left[ {{\left( 1\ +\ \frac{\Delta T}{{{T}_{0}}} \right)}^{\text{4}}}-1 \right]$

Using binomial approximation, $\displaystyle {{\left( 1\ +\ \frac{\Delta T}{{{T}_{0}}} \right)}^{\text{4}}}=1+\text{4}\frac{\Delta T}{{{T}_{0}}}$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\frac{\sigma A}{ms}T_{0}^{4}\left[ \left( 1\ +\ 4\frac{\Delta T}{{{T}_{0}}} \right)-1 \right]$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\left( \frac{4\sigma A}{ms}T_{0}^{3} \right)\Delta T=\left( \frac{4\sigma A}{ms}T_{0}^{3} \right)\left( T-{{T}_{0}}\ \right)$

$\displaystyle \therefore \frac{d\theta }{dt}=K\left( T-{{T}_{0}}\ \right)$

जहाँ $\displaystyle K=\left( \frac{4\sigma A}{ms}T_{0}^{3} \right)$ .

Solving the Differential Equation for Newton’s Law of Cooling

Using equation (3) of Newton’s Law of Cooling :

$\displaystyle -\frac{dT}{dt}=K(T-{{T}_{0}})$

− sign → temp falls

$\displaystyle \Rightarrow \frac{dT}{(T-{{T}_{0}})}=-Kdt$

Integrating both sides,

$\displaystyle \int\limits_{{{T}_{i}}}^{{{T}_{f}}}{\frac{dT}{(T-{{T}_{0}})}}=-K\int\limits_{0}^{t}{dt}$

where T_i = initial temperature of the object and T_f = final temperature of the object.

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}(T-{{T}_{0}}) \right]_{{{T}_{i}}}^{{{T}_{f}}}=-K\left( t \right)_{0}^{t}$

$\displaystyle \Rightarrow lo{{g}_{e}}\left[ \frac{{{T}_{f}}-{{T}_{0}}}{{{T}_{i}}-{{T}_{0}}} \right]=-Kt$

$\displaystyle \Rightarrow \frac{{{T}_{f}}-{{T}_{0}}}{{{T}_{i}}-{{T}_{0}}}={{e}^{-Kt}}$

$\displaystyle \Rightarrow \left( {{T}_{f}}-{{T}_{0}} \right)=\left( {{T}_{i}}-{{T}_{0}} \right){{e}^{-Kt}}$

$\displaystyle \Rightarrow {{T}_{f}}={{T}_{0}}+\left( {{T}_{i}}-{{T}_{0}} \right){{e}^{-Kt}}$ …..(8)

Example 1.

A body cools in 5 minutes from 64 °C to 52 °C. What will be its temperature after the next 5 minutes ? The temperature of the surroundings is 10 °C.

Solution :

Method – I

According to Newton’s Law of Cooling,

$\displaystyle \left[ \frac{{{T}_{1}}-{{T}_{2}}}{t} \right]=K\left[ \frac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{0}} \right]$

$\displaystyle \Rightarrow \left[ \frac{64-52}{5} \right]=K\left[ \frac{64+52}{2}-10 \right]$

$\displaystyle \Rightarrow K=\frac{1}{20}$

After 5 mintutes,

$\displaystyle \left[ \frac{52-{{T}_{2}}}{5} \right]=\left( \frac{1}{20} \right)\times \left[ \frac{52+{{T}_{2}}}{2}-10 \right]$

$\displaystyle \Rightarrow 8\times \left( 52-{{T}_{2}} \right)=32+{{T}_{2}}$

$\displaystyle \therefore {{T}_{2}}=42.67{}^\circ C$

Method – II

From equation (8),

$\displaystyle {{T}_{f}}={{T}_{0}}+\left( {{T}_{i}}-{{T}_{0}} \right){{e}^{-Kt}}$

$\displaystyle \Rightarrow \frac{\left( {{T}_{f}}-{{T}_{0}} \right)}{\left( {{T}_{i}}-{{T}_{0}} \right)}={{e}^{-Kt}}$

$\displaystyle \Rightarrow \frac{\left( {{T}_{i}}-{{T}_{0}} \right)}{\left( {{T}_{f}}-{{T}_{0}} \right)}={{e}^{Kt}}$

$\displaystyle \therefore t=\frac{1}{K}lo{{g}_{e}}\left[ \frac{\left( {{T}_{i}}-{{T}_{0}} \right)}{\left( {{T}_{f}}-{{T}_{0}} \right)} \right]$

Now

$\displaystyle 5=\frac{1}{K}lo{{g}_{e}}\left[ \frac{\left( 64-10 \right)}{\left( 52-10 \right)} \right]$

$\displaystyle \Rightarrow 5=\frac{1}{K}lo{{g}_{e}}\left[ \frac{9}{7} \right]$ …..(a)

After 5 mintutes,

$\displaystyle 5=\frac{1}{K}lo{{g}_{e}}\left[ \frac{\left( 52-10 \right)}{\left( {{T}_{f}}-10 \right)} \right]$

$\displaystyle \Rightarrow 5=\frac{1}{K}lo{{g}_{e}}\left[ \frac{42}{\left( {{T}_{f}}-10 \right)} \right]$ …..(b)

Now using equation (a) and (b),

$\displaystyle \frac{1}{K}lo{{g}_{e}}\left[ \frac{9}{7} \right]=\frac{1}{K}lo{{g}_{e}}\left[ \frac{42}{\left( {{T}_{f}}-10 \right)} \right]$

$\displaystyle \Rightarrow \left[ \frac{9}{7} \right]=\left[ \frac{42}{\left( {{T}_{f}}-10 \right)} \right]$

$\displaystyle \therefore {{T}_{f}}={{42.67}^{{}^\circ }}C$

Example 2.

If a liquid takes 30 seconds in cooling from 80°C to 70°C and 70 seconds in cooling from 60°C to 50°C, then find the room temperature.

Solution :

In first case :

$\displaystyle \frac{80-70}{30}=K\left( \frac{80+70}{2}-{{T}_{0}} \right)$

$\displaystyle \Rightarrow \frac{1}{3}=K\left( 75-{{T}_{0}} \right)$ …..(a)

In second case :

$\displaystyle \frac{60-50}{70}=K\left( \frac{60+50}{2}-{{T}_{0}} \right)$

$\displaystyle \Rightarrow \frac{1}{7}=K\left( 55-{{T}_{0}} \right)$ …..(b)

Dividing equation (a) by equation (b), we get

$\displaystyle \frac{7}{3}=\frac{75-{{T}_{0}}}{55-{{T}_{0}}}$

$\displaystyle \Rightarrow {{T}_{0}}=40{}^\circ C$

Example 3.

A liquid takes 30 seconds to cool from 95 ºC to 90 ºC and 70 seconds to cool from 55 ºC to 50 ºC. Find the room temperature and the time it will take to cool from 50 ºC to 45 ºC.

Solution :

From the first data :

$\displaystyle \left[ \frac{95-90}{30} \right]=K\left[ \frac{95+90}{2}-{{T}_{0}} \right]$

$\displaystyle \Rightarrow \frac{1}{6}=K\left[ \frac{185}{2}-{{T}_{0}} \right]$ …..(a)

From the second data :

$\displaystyle \left[ \frac{55-50}{70} \right]=K\left[ \frac{55+50}{2}-{{T}_{0}} \right]$

$\displaystyle \Rightarrow \frac{1}{14}=K\left[ \frac{105}{2}-{{T}_{0}} \right]$ …..(b)

Dividing equation (a) by (b), we get

$\displaystyle \frac{7}{3}=\frac{\frac{185}{2}-{{T}_{0}}}{\frac{105}{2}-{{T}_{0}}}$

$\displaystyle \Rightarrow \frac{735}{2}-7{{T}_{0}}=\frac{555}{2}-3{{T}_{0}}$

$\displaystyle \Rightarrow \frac{735}{2}-\frac{555}{2}=4{{T}_{0}}$

$\displaystyle \Rightarrow 367.5-277.5=4{{T}_{0}}$

$\displaystyle \Rightarrow {{T}_{0}}={{22.5}^{{}^\circ }}C$

Let the time taken in cooling from 50 ºC to 45 ºC is t sec, then

$\displaystyle \left[ \frac{50-45}{t} \right]=K\left[ \frac{50+45}{2}-22.5 \right]$

$\displaystyle \Rightarrow \frac{1}{t}=5K$ …..(c)

Dividing equation (a) by (c), we get

$\displaystyle \frac{t}{6}=\frac{\left[ \frac{185}{2}-22.5 \right]}{5}$

$\displaystyle \therefore t=84sec$

To determine the specific heat of a given liquid using Newton’s Law of Cooling

Let equal volumes of two liquids having densities ρ_₁, ρ_₂ and specific heats s_₁, s_₂ respectively are placed in calorimeters of the same surface area. Both liquids are allowed to cool from the same initial temperature T₁ to the same final temperature T₂ in the same surrounding temperature T₀ . The cooling takes place in time intervals t₁ and t₂ respectively. If the water equivalent of each calorimeter is w, then according to Newton’s law of cooling :

$\displaystyle {{\left( \frac{dQ}{dt} \right)}_{water}}={{\left( \frac{dQ}{dt} \right)}_{liquid}}$

$\displaystyle \Rightarrow \frac{\left( w+{{m}_{1}}{{s}_{1}} \right)\left( {{T}_{1}}-{{T}_{2}} \right)}{{{t}_{1}}}=\frac{\left( w+{{m}_{2}}{{s}_{2}} \right)\left( {{T}_{1}}-{{T}_{2}} \right)}{{{t}_{2}}}$

$\displaystyle \therefore \frac{\left( w+{{m}_{1}}{{s}_{1}} \right)}{{{t}_{1}}}=\frac{\left( w+{{m}_{2}}{{s}_{2}} \right)}{{{t}_{2}}}$ …..(9)

Note :

If water equivalent w of calorimeter is negligible then,

$\displaystyle \frac{{{m}_{1}}{{s}_{1}}}{{{t}_{1}}}=\frac{{{m}_{2}}{{s}_{2}}}{{{t}_{2}}}$

$\displaystyle \Rightarrow \frac{\left( V{{\rho }_{1}} \right){{s}_{1}}}{{{t}_{1}}}=\frac{\left( V{{\rho }_{2}} \right){{s}_{2}}}{{{t}_{2}}}$

$\displaystyle \Rightarrow \frac{{{\rho }_{1}}{{s}_{1}}}{{{\rho }_{2}}{{s}_{2}}}=\frac{{{t}_{1}}}{{{t}_{2}}}$ …..(10)

With the help of equations (9) and (10), we can find the specific heat of given liquid.

Example 4.

A calorimeter contains 40 g of water at 50°C and the temperature falls to 45 °C in 10 minutes. When the same calorimeter contains 100g of water at 50 °C, it takes 20 minutes for the temperature to become 45 °C. Find the water equivalent of the calorimeter.

Solution :

$\displaystyle \frac{\left( w+{{m}_{1}}{{s}_{1}} \right)}{{{t}_{1}}}=\frac{\left( w+{{m}_{2}}{{s}_{2}} \right)}{{{t}_{2}}}$

$\displaystyle \Rightarrow \frac{\left( w+40\times 1 \right)}{10}=\frac{\left( w+100\times 1 \right)}{20}$

$\displaystyle \Rightarrow 2w+80=w+100$

$\displaystyle \therefore w=20gram$

Example 5.

Two liquids of same volume are cooled under same conditions from 65 ºC to 50 ºC. Time taken are 200 sec and 480 sec. If ratio of their specific heats is 2 : 3 then find the ratio of their densities. (neglect the water equivalent of calorimeter)

Solution :

Here w₁ = w₂ = 0, hence using

$\displaystyle \frac{{{\rho }_{1}}{{s}_{1}}}{{{\rho }_{2}}{{s}_{2}}}=\frac{{{t}_{1}}}{{{t}_{2}}}$

$\displaystyle \Rightarrow \frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{{{t}_{1}}}{{{t}_{2}}}\times \frac{{{s}_{2}}}{{{s}_{1}}}$

$\displaystyle \Rightarrow \frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{200}{480}\times \frac{3}{2}$

$\displaystyle \therefore \frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{5}{8}$

Example 6.

A calorimeter of water equivalent 5 × 10^-3 kg contains 25 × 10^-3 kg of water. It takes 3 minutes to cool from 28 °C to 21 °C. When the same calorimeter is filled with 30 × 10^-3 kg of turpentine oil then it takes 2 minutes to cool from 28° C to 21 °C. Find out the specific heat of turpentine oil.

Solution :

$\displaystyle {{\left( \frac{dQ}{dt} \right)}_{water}}={{\left( \frac{dQ}{dt} \right)}_{turpentine}}$

$\displaystyle \Rightarrow \frac{\left( w+{{m}_{1}}{{s}_{1}} \right)}{{{t}_{1}}}=\frac{\left( w+{{m}_{2}}{{s}_{2}} \right)}{{{t}_{2}}}$

$\displaystyle \Rightarrow \frac{\left( 5\times {{10}^{-3}}+25\times {{10}^{-3}}\times 1 \right)}{3}=\frac{\left( 5\times {{10}^{-3}}+30\times {{10}^{-3}}\times {{s}_{2}} \right)}{2}$

$\displaystyle \therefore {{s}_{2}}=0.5cal.{{g}^{-1}}{}^\circ {{C}^{-1}}$

$\displaystyle {{s}_{2}}=0.5kcal.k{{g}^{-1}}{}^\circ {{C}^{-1}}$

Example 7.

An ice cube at temperature $- 2 0 °C$ is kept in a room at temperature $2 0 °C$ . The variation of temperature of the body with time is given by :

Solution :

(1). From −20 °C to 0 °C

Ice temperature increases due to heat from the room.
Newton’s law of cooling/heating applies :

$\displaystyle \frac{dT}{dt}\propto ({{T}_{room}}-T)$

so as rises, (becomes smaller → rate of rise decreases.

Graph must be concave downward (slope starts high, then gets smaller as it approaches 0 °C).

(2). At 0 °C (Melting)

Temperature stays constant during phase change until all ice melts.
Graph should have a horizontal plateau at .

(3). From 0 °C to 20 °C

Now we have liquid water. Again Newton’s law applies.
Curve is concave downward approaching asymptotically.

Now from the given graphs :

(1) → First portion is concave upward (slope increasing), which means temperature rises faster and faster as it approaches 0 °C.

This is physically wrong — warming should slow down as .

(2) → First portion is concave downward (slope decreasing), then flat at 0 °C, then concave downward again.

This exactly matches what we expect. ✅

(3) → Shows cooling (temperature decreasing), not heating. ❌

(4) → Shows weird wavy behaviour, no reason for that physically. ❌

Hence Graph (2) is correct for this situation.

Newton’s Law of Cooling

Limitations of Newton’s Law of Cooling

Derivation of Newton’s Law of Cooling from Steafan’s Boltzman Law

Solving the Differential Equation for Newton’s Law of Cooling

To determine the specific heat of a given liquid using Newton’s Law of Cooling

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