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Newton's Law Of Cooling MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

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Variation of curve's for Newton's Law of Cooling is considered one of the most asked concept.
22 Questions around this concept.

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EASY

Two identical beakers A and B contain equal volumes of two different liquids at $60^{\circ} \mathrm{C}$ each and are left to cool down. The liquid in A has a density of $8 \times 10^2 \mathrm{~kg} / \mathrm{m}^3$ and specific heat of $2000 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ while the liquid in B has a density of $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and specific heat of $4000 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

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A liquid in a beaker has temperature $\Theta(t)$ at time $t$ and $\Theta_0$ is the temperature of surroundings, then according to Newton's law of cooling the correct graph between $\log _e\left(\Theta-\Theta_0\right)$ and t is :

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Concepts Covered - 1

Newton's law of cooling

Rate of Loss of Heat-

If an ordinary body at temperature $\theta$ is placed in an environment of temperature $\theta_{0}( \ and \ \ \theta_{0}<\theta)$ then heat loss by radiation.

And Rate of Loss of Heat is given by

from Stefan Boltzmann law

$R_H=\frac{dQ}{dt}=A\epsilon\sigma(\theta^{4}-\theta_{0}^{4})$ .....(1)

Rate of Cooling-

If m is the body and c is the specific heat then $Q=mc\Delta \theta$

And $\frac{dQ}{dt}=mc \frac{\Delta \theta }{dt}$ .....(2)

Comparing equation 1 and 2

we get Rate of Cooling as

$R_{c}=\frac{\Delta \theta }{dt}=\frac{A\epsilon\sigma}{mc}(\theta^{4}-\theta_{0}^{4})$

where

c = specific heat capacity

R_c= Rate of cooling.

As $mass=volume*density\Rightarrow \ m=\rho v$

So We can also write

$R_{c}=\frac{A\epsilon\sigma}{v\rho c}(\theta^{4}-\theta_{0}^{4})$

If two bodies of the same material under identical environments

$\frac{(R_{c})_{1}}{(R_{c})_2}=\frac{A_{1}v_{2}}{A_{2}v_{1}}$

Dependence of the rate of cooling ( $R_{c}$ )

As $R_{c} =\frac{A\epsilon\sigma}{mc}(\theta^{4}-\theta_{0}^{4})=\frac{A\epsilon\sigma}{v\rho c}(\theta^{4}-\theta_{0}^{4})$

So $R_{c}$ will depend on

Nature of radiating surface i.e. greater the emissivity faster will be the cooling.
Area of the radiating surface, i.e. greater the area of the radiating surface, faster will be the cooling.
Specific heat of radiating body i.e. greater the specific heat of radiating body slower will be cooling.
Mass of radiating body i.e. greater the mass of radiating body slower will be the cooling.
The temperature of the radiating body i.e. greater the temperature of the body faster will be cooling.
The temperature of surrounding i.e. greater the temperature of surrounding slower will be cooling.

Newton's Law of Cooling

According to Newton's Law of Cooling, if the temperature difference between the body and its surrounding is very small then the Rate of cooling is directly proportional to the temperature difference between the body and its surrounding.

I.e $\frac{d\theta}{dt}\alpha(\theta-\theta_{0})$

Greater the temperature difference between the body and its surrounding greater will be the rate of cooling.
$if \ \theta =\theta _0\Rightarrow \frac{d\theta }{dt}=0$ i.e. a body can never be cooled to a temperature lesser than its surrounding by radiation.
When body Cools by Radiation from $\ \theta_1^0 C$ to theta $\ \theta_2^0 C$ in time t Then $\left[\frac{\theta_{1}-\theta_{2}}{t} \right ]=k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0} \right ]$ Where $\theta_{av}=\frac{\theta_{1}+\theta_{2}}{2}$