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

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\\)

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 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 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\\)

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\\)

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 a temperature \\(3T\\) to \\(2T\\) in \\(10\\) minutes. The room temperature is \\(T\\). Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next \\(10\\) minutes will be

From newton's law of cooling,\\(ln(\dfrac{3T-T}{2T-T})=kt\\)\\(=ln(2)=k(10)\\)For next 10 minutes,\\(ln(\dfrac{2T-T}{T'-T})=k(10)\\)\\(\implies \dfrac{2T-T}{T'-T}=2\\)\\(\implies T'=\dfrac{3}{2}T\\)

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

Q: A body cools from a temperature \\(3T\\) to \\(2T\\) in \\(10\\) minutes. The room temperature is \\(T\\). Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next \\(10\\) minutes will be

From newton's law of cooling,\\(ln(\dfrac{3T-T}{2T-T})=kt\\)\\(=ln(2)=k(10)\\)For next 10 minutes,\\(ln(\dfrac{2T-T}{T'-T})=k(10)\\)\\(\implies \dfrac{2T-T}{T'-T}=2\\)\\(\implies T'=\dfrac{3}{2}T\\)

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