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>> $Newton's Law of Cooling$

Newton's Law of Cooling

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Newton's Law of Cooling

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Newton's Law of Cooling - L1

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Q: A body cools from \${80}^{\circ}\$C to \${50}^{\circ}\$C in \$5\$ minutes. Calculate the time it takes to cool from \${60}^{\circ}\$C to \${30}^{\circ}\$C. The temperature of the surroundings is \${20}^{\circ}\$C.

Here, Initial temperature \$( {T}_{i}) = {80}^{\circ}\$C Final temperature \$( {T}_{f}) = {50}^{\circ}\$C Temperature of the surrounding \$( {T}_{0}) = {20}^{\circ}\$C \$ t = 5\$ min According to Newton's law of cooling,Rate of cooling \$\dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}\right)}{2}-{T}_{o}]\$\$\dfrac{\left({T}_{f}-{T}_{i}\right)}{t}=K\left[\dfrac{\left(80 +50\right)}{2} - 20\right]\$\$\dfrac{80-50}{5} = K[ 65 - 20]\$\$6 = K\times 45\$\$K = \dfrac{6}{45} = \dfrac{2}{15}\$ In second condition, initial temperature \$={T}_{i}= {60}^{\circ}\$C Final temperature \${T}_{f}={30}^{\circ}\$C Time taken for cooling is \$t\$ According to Newton's law of cooling \$\dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30\right)}{2} -20\right]\$\$\dfrac{30}{t}=\dfrac{2}{15}\times 25\$\$\dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3}\$\$\therefore t = 9\$ min

Q: A body cools down from \$80^o C\$ to \$60^oC\$ in \$10\$ Minutes when the temperature of surrounding is \$30^oC\$. The temperature of the body after next \$10\$ minutes will be:

\$\cfrac{dT}{dt}=-K\left(T-T_s\right)+T_in\$\$\therefore\$ Here from newtons law of co0ling\$\cfrac{dT}{dt}=-k\left(\cfrac{T_f+T_i}{2}-T_s\right)\$\$\therefore \cfrac{T_f-T_i}{\triangle t}=K\left[\cfrac{T_f+T_i}{2}-T_s\right]\$\$\therefore K=\cfrac{1}{20}\$\$\Rightarrow=\cfrac{60-T}{10}=\cfrac{1}{20}\left[\cfrac{60+T}{2}-30\right]\$\$=120-2T=\cfrac{T+60}{21}-30\$\$\cfrac{5T}{2}=120\$\$T=48\$

Q: Certain quantity of water cools from \$70^oC\$ to \$60^oC\$ in the first \$5\$ minutes and to \$54^oC\$ in the next \$5\$ minutes. The temperature of the surroundings is ;

\$\dfrac {60-70}{5}=-K(65-T)\$\$\dfrac {54-60}{5}-K(57-T)\$\$\dfrac {-10}{-6}=\dfrac {65-T}{57-T}\$\$285-5T=195-3T\$\$90=2T\$\$T=45^o\$

Q: A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

The circular plate has maximum surface area and as such it cools the fastest. For a given volume (mass in this case),a sphere has the least surface area and accordingly cools the slowest.

Q: A body cools in \$ 7\$ minutes from \$60^o C\$ to \$40^o C\$. If the temperature of surrounding is \$10^o C\$, the temperature after next \$7\$ minutes will be :

From newton's law of cooling\$\dfrac {dT}{dt}=k(T-T_{surrounding})\$...(i)integrating equation (i)\$\dfrac {T_f-T_{surrounding}}{T_i-T_{surrounding}}=e^{-kt}\$..(ii)From given data in question in first \$7\$ minutes \$T_f=40 \;C;\;T_i=60\; C;\; T_{surrounding}= 10^o \;C\$\$ \displaystyle \frac {40-10}{60-10} =e^{-k7} , \$...(iii)Now the temperature decrease further due to temperature gradientFrom given data in question in next \$7\$ minutes \$T_f=?;\;T_i=40\; C;\; T_{surrounding}= 10^o \;C\$applying equation (i)\$ \displaystyle \frac {T_f-10}{40-10} =e^{-k7} , \$...(iv)substituting value of \$e^{-k7} , \$ from equation(iii)\$\dfrac{T_f-10}{40-10} =\dfrac{3}{5} \$\$ \displaystyle T_f = 10+18=28^o \;C\$Hence correct option is D

Q: A body cools from 50\$^{o}\$C to 45\$^{o}\$C in 5 min and to 40\$^{o}\$C in another 8 min. The temperature of the surrounding is

Newton's cooling law \$\dfrac{d\theta}{dt}=-k(\theta-{\theta}_{0})\$putting the given values of the situation in the formula we get:-\$ \dfrac{(50-45)}{5} = -k (50-T)\$\$=> 1 = -k (50-T) \$ -----(1)Now, cooling body at takes 8 min to reach \$40{}^{0}C\$ form \$45{}^{0}C\$\$\dfrac{(45-40)}{8} = -k (45-T)\$\$\dfrac{5}{8} = -k (45-T)\$ ------(2)Dividing the two equations, we get:\$8(45-T) = 5(50-T)\$\$360-8T=250-5T\$\$T=110/3=36.67^0C\$

Q: 100 gm of ice at \$0^{_-^0}C\$ is mixed with 100 g of water at \$100^{_-^0} C\$. What will be the final temperature of the mixture?

\$\theta\$ is temperature.\$\theta_{mix}\$ is given by : \$=\dfrac{\theta_W - \dfrac{L_i}{c_W}}{2}\$putting values:\$\theta_{mix} = \dfrac{\theta_W - \dfrac{L_i}{c_W}}{2} = \dfrac{100 - \dfrac{80}{1}}{2} = 10^\circ C\$

Q: Hot water cools from \$60^oC\$ to \$50^oC\$ in the first \$10\$ minutes and to \$42^oC\$ in the next \$10\$ minutes. The temperature of the surroundings is :

Answer is BWe know from Newton's law of cooling that .\$T= T_f + (T_i - T_f)e^{-kt}\$\$50= T_f+(60-T_f)e^{-10k}\$\$50= y+60x \$ _______________(i) \$ where \ x= e^{-10k} \$and \$y= T_f(1-e^{-10k}) \$Also,\$42= T_f+(50-T_f)e^{-10k}\$\$42= y+50x \$ ______________ (ii) solving (i) and (ii), we get\$x= e^{-10k}= 4/5\$and,\$y=T_f(1-e^{-10k})= 2\$\$T_f= 10^{o}C\$

Q: A pendulum clock is \$5\$sec fast at temperature of \$15^oC\$ and \$10\$sec slow at a temperature of \$30^oC\$. At temperature does it give the correct time-

\$T=2\pi\sqrt{\dfrac{l}{g}}\$\$\Rightarrow T\propto l^{1/2}\$\$\Rightarrow \dfrac{\Delta T}{T}=\dfrac{1}{2}\dfrac{\Delta l}{l}\$and \$\Delta l=l\propto \Delta t\$, linear expansion formula\$\Rightarrow \dfrac{\Delta l}{l}=\propto \Delta t\$\$\Delta t=\$ change in temperature\$=\$Final\$-\$initial\$5\$ sec fast\$\Rightarrow 1\$ hour alarm rises \$5\$ sec before i.e., time period reduced\$\Rightarrow \Delta T=-5\$Case I:\$5\$ sec fast at \$15^o\$C\$\dfrac{-5}{T}=\dfrac{1}{z}\propto \Delta t=\dfrac{1}{2}\propto (15-t)\$ ……..\$(1)\$\$t=\$ temperature at which correct reading.\$15^o < t < 30^o\$Case II:\$10\$ sec slow at \$30^o\$C\$\Rightarrow \Delta T=10\$sec\$\Rightarrow \dfrac{10}{T}=\dfrac{1}{2}\propto (30-t)\$ ………\$(2)\$Heated to \$30^o\$Cdividing \$(1)\$ by \$(2)\$We get\$-0.5=\dfrac{15-t}{30-t}\$\$\Rightarrow 30-t=2(t-15)\$\$\Rightarrow t=20^oC\$.

Newton's Law of Cooling I

14 mins

Newton's Law of Cooling - II

4 mins

Experimental Setup of Verification of Newton's Law of Cooling

15 mins

Quick Summary With Stories

Newton's Law of Cooling

3 mins read

Experimental Setup of Verification of Newton's Law of Cooling

3 mins read

Problem based on Newton's law cooling

2 mins read

Important Questions

A body cools from $80^{\circ}$ C to $50^{\circ}$ C in $5$ minutes. Calculate the time it takes to cool from $60^{\circ}$ C to $30^{\circ}$ C. The temperature of the surroundings is $20^{\circ}$ C.

Medium

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A body cools down from $8 0^{o} C$ to $6 0^{o} C$ in $10$ Minutes when the temperature of surrounding is $3 0^{o} C$ . The temperature of the body after next $10$ minutes will be:

Medium

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Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;

Medium

NEET

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A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

Medium

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A body cools in $7$ minutes from $6 0^{o} C$ to $4 0^{o} C$ . If the temperature of surrounding is $1 0^{o} C$ , the temperature after next $7$ minutes will be :

Medium

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A body cools from 50 $^{o}$ C to 45 $^{o}$ C in 5 min and to 40 $^{o}$ C in another 8 min. The temperature of the surrounding is

Hard

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A body cools from $6 0^{\circ} C$ to $5 0^{\circ} C$ in $10$ minutes. If the room temperature is $2 5^{\circ} C$ and assuming Newton's cooling law holds good, the temperature of the body at the end of the next $10$ minutes is:

Medium

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100 gm of ice at $0^{_{-}^{0}} C$ is mixed with 100 g of water at $10 0^{_{-}^{0}} C$ . What will be the final temperature of the mixture?

Medium

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Hot water cools from $6 0^{o} C$ to $5 0^{o} C$ in the first $10$ minutes and to $4 2^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

Medium

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A pendulum clock is $5$ sec fast at temperature of $1 5^{o} C$ and $10$ sec slow at a temperature of $3 0^{o} C$ . At temperature does it give the correct time-

Medium

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Revise with Concepts

Newton's Law of Cooling

Newton's Law of Cooling - L1

Learn with Videos

Quick Summary With Stories

Newton's Law of Cooling

Experimental Setup of Verification of Newton's Law of Cooling

Problem based on Newton's law cooling

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.

A body cools down from 80oC to 60oC in 10 Minutes when the temperature of surrounding is 30oC. The temperature of the body after next 10 minutes will be:

Certain quantity of water cools from 70oC to 60oC in the first 5 minutes and to 54oC in the next 5 minutes. The temperature of the surroundings is ;

A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

A body cools in 7 minutes from 60oC to 40oC. If the temperature of surrounding is 10oC, the temperature after next 7 minutes will be :

A body cools from 50oC to 45oC in 5 min and to 40oC in another 8 min. The temperature of the surrounding is

A body cools from 60∘C to 50∘C in 10 minutes. If the room temperature is 25∘C and assuming Newton's cooling law holds good, the temperature of the body at the end of the next 10 minutes is:

100 gm of ice at 0−0​C is mixed with 100 g of water at 100−0​C. What will be the final temperature of the mixture?

Hot water cools from 60oC to 50oC in the first 10 minutes and to 42oC in the next 10 minutes. The temperature of the surroundings is :

A pendulum clock is 5sec fast at temperature of 15oC and 10sec slow at a temperature of 30oC. At temperature does it give the correct time-

A body cools from $80^{\circ}$ C to $50^{\circ}$ C in $5$ minutes. Calculate the time it takes to cool from $60^{\circ}$ C to $30^{\circ}$ C. The temperature of the surroundings is $20^{\circ}$ C.

A body cools down from $8 0^{o} C$ to $6 0^{o} C$ in $10$ Minutes when the temperature of surrounding is $3 0^{o} C$ . The temperature of the body after next $10$ minutes will be:

Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;

A body cools in $7$ minutes from $6 0^{o} C$ to $4 0^{o} C$ . If the temperature of surrounding is $1 0^{o} C$ , the temperature after next $7$ minutes will be :

A body cools from 50 $^{o}$ C to 45 $^{o}$ C in 5 min and to 40 $^{o}$ C in another 8 min. The temperature of the surrounding is

A body cools from $6 0^{\circ} C$ to $5 0^{\circ} C$ in $10$ minutes. If the room temperature is $2 5^{\circ} C$ and assuming Newton's cooling law holds good, the temperature of the body at the end of the next $10$ minutes is:

100 gm of ice at $0^{_{-}^{0}} C$ is mixed with 100 g of water at $10 0^{_{-}^{0}} C$ . What will be the final temperature of the mixture?

Hot water cools from $6 0^{o} C$ to $5 0^{o} C$ in the first $10$ minutes and to $4 2^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

A pendulum clock is $5$ sec fast at temperature of $1 5^{o} C$ and $10$ sec slow at a temperature of $3 0^{o} C$ . At temperature does it give the correct time-