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

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

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

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.

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

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 .

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

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

Spheres P and Q are uniformly constructed from the same material which is a good conductor of heat and the radius of Q is thrice the radius of P . The rate of fall of temperature of P is x times that of Q . When both are at the same surface temperature, the value of x is :

From Newton's law of cooling, \displaystyle \frac{dT}{dt}=-\frac{4\sigma eAT_0^3}{ms}(T-T_0)=-\frac{4\sigma e(4\pi r^2)T_0^3}{(\rho 4/3\pi r^3)s}(T-T_0) Thus, \displaystyle \frac{dT}{dt} \propto \frac{1}{r} \displaystyle \frac{\left(\frac{dT}{dt}\right)_P}{\left(\frac{dT}{dt}\right)_Q}=\frac{r_Q}{r_P}=\frac{3r_P}{r_P}=3 \displaystyle \therefore \left(\frac{dT}{dt}\right)_P=3\times \left(\frac{dT}{dt}\right)_Q

Two identical masses of 5\ kg each fall on a wheel from a height of 10\ m . The wheel disturbs a mass of 2\ kg water, the rise in temperature of water will be:

Given, m=5kg M=2kg h=10m g=10m/s^2 J=4.2joule We know that, W=JQ 2mgh=JMs\Delta T \Delta T=\dfrac{2mgh}{JMs} \Delta T=\dfrac{2\times 5\times 10\times 10}{4.2\times 2\times 1000} \Delta T=0.12^0C The correct option is D.

Newton's Law of Cooling | Definition, Examples, Diagrams

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

diagram

Plot of temperature vs time for Newton's Law of cooling

diagram

Plot of log of temperature difference vs time for Newton's Law of cooling

definition

Experimental verification of Newton's Law of cooling

Newton's Law of cooling can be verified with the experimental set-up shown in the figure. The set-up consists of a double walled vessel (V) containing water in between the walls. A copper calorimeter (C) containing hot water is placed inside the double walled vessel. Two thermometers through the corks are used to measure the temperatures

T_{2}

of water in calorimeter and

T_{1}

of hot water in between the double walls respectively. Temperature of hot water in the calorimeter is noted after fixed intervals of time. A graph is plotted between

l o g (T_{2} - T_{1})

and time(t). The nature of graph is observed to be a straight line having a negative slope which is in accordance of the equation of Newton's Law of cooling.

definition

Describe the limitations of Newton's Law of Cooling

The limitations of Newton's Law of cooling are as follows:
1. It can be used only at lower temperatures.
2. It can only be use when difference between initial and final temperatures is less than 50-80 degree Celsius.
3. It holds true only if the temperature of the surroundings remains constant throughout the cooling of the body.

definition