Heat Transfer Concept Page - 9

Example

Problems using Newton's Law of cooling

A body cools from 70o${}^o$C to 50o${}^o$C in 5 minutes. Temperature of surroundings is 20o${}^o$C. Its temperature after next 10 minutes is:Newton's cooling law  dθdt=−k(θ−θ0)${\displaystyle \frac{d\theta }{dt}}=-k(\theta -{\theta }_0)$
putting the given values of the situation in the formula we get:-
(70−50)5=−k(70−20)${\displaystyle \frac{(70-50)}{5}}=-k(70-20)$
=>205=−k(50)$=>{\displaystyle \frac{20}{5}}=-k(50)$
=>−k=225$=>-k={\displaystyle \frac{2}{25}}$
now, cooling body at 500C$50{}^{0}C$ for 10 min
let temperature after 10 min be T 
(50−T)10=225(50−20)${\displaystyle \frac{(50-T)}{10}}={\displaystyle \frac{2}{25}}(50-20)$
(50−T)10=125${\displaystyle \frac{(50-T)}{10}}={\displaystyle \frac{12}{5}}$
50−T=24$50-T=24$
T=50−24$T=50-24$
T=260C$T={26}^{0}C$

Example

Example of newton's law of cooling in phase change

Example:
A spherical iron ball 10$10$ cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3/min$50c{m}^3/min$. When the thickness of ice is 5 cm$5cm$, then find the rate at which the thickness of ice decreases. Solution:
Let inner radius of ice layer = Ri$R_i$ and outer radius = Ro$R_o$.Therefore thickness = Ro−Ri$R_o-R_i$.
During melting, only the outer radius decreases whereas the inner radius is unchanged. So the thickness changes only due to the change in outer radius.
∴ddt(Ro−Ri)=dRodt$\therefore {\displaystyle \frac{d}{dt}}(R_o-R_i)={\displaystyle \frac{d{R}_o}{dt}}$.Volume of ice = V=4π3(R3o−R3i)$V={\displaystyle \frac{4\pi }{3}}(R_o^3-R_i^3)$ ⇒dVdt=4πR2odRodt=50cm3/min$\Rightarrow {\displaystyle \frac{dV}{dt}}=4\pi R_o^2{\displaystyle \frac{d{R}_o}{dt}}=50c{m}^3/min$
When thickness of layer is 5cm$5cm$, Ro=15cm$R_o=15cm$.
So, dRodt=14πR2odVdt=50900π=118πcm/min${\displaystyle \frac{d{R}_o}{dt}}={\displaystyle \frac{1}{4\pi R_o^2}}{\displaystyle \frac{dV}{dt}}={\displaystyle \frac{50}{900\pi }}={\displaystyle \frac{1}{18\pi }}cm/min$.
So, rate of decrease of thickness is 118πcm/min${\displaystyle \frac{1}{18\pi }}cm/min$.

Definition

Regelation

Take a slab of ice. Take a wire and fix two blocks,say 5 kg each, at its ends. Put the wire over the slab. You will observe that the wire passes through the ice slab. This happens due to the fact that just below the wire, ice melts at lower temperature due to increase in pressure. When the wire has passed, water above the wire freezes again. Thus the wire passes through the slab and the slab doesn't split. This process of refreezing is called regelation. 

Definition

Factors affecting melting point

Parameters which affect melting point are:

• Ionic Bonds
• Intermolecular Forces
• Shape of Molecules
• Size of Molecule

Result

Effect of pressure on melting point

Melting point of substances that contract on melting (like ice) decreases with increase in pressure. 
Melting point of substances that expand on melting (like lead) increases with increase in pressure. 

Result

Effect of impurities on melting point

Melting point of a  substance decreases by the presence of impurities in it. This phenomenon is used in making the freezing mixture by adding salt to ice.

Definition

Humidity, dew point and fog

Humidity is the amount of water vapor in the air. Water vapor is the gaseous state of water and is invisible.There are three main measurements of humidity: absolute, relative and specific. Absolute humidity is the water content of air at a given temperature expressed in gram per cubic meter.Relative humidity, expressed as a percent, measures the current absolute humidity relative to the maximum (highest point) for that temperature. Specific humidity is a ratio of the water vapor content of the mixture to the total air content on a mass basis.
Dew point is the highest temperature at which airborne water vapor will condense to form liquid dew. A higher dew point means there will be more moisture in the air.
Fog is a visible mass consisting of cloud water droplets or ice crystals suspended in the air at or near the Earth's surface. Fog can be considered a type of low-lying cloud and is heavily influenced by nearby bodies of water, topography, and wind conditions.

Example

calculation of Relative Humidity

Q. If the actual vapor density is 10  g/m3$10g/m^3$ at 20oC${20}^{o}C$ while the saturation vapor density at the same temperature is 17.3  g/m3$17.3g/m^3$. Then find the relative humidity.
Solution :
Step 1 : Formula for Relative Humidity is given as
                                   Relative Humidity=actual vapor densitysaturation vapor density×100%$RelativeHumidity={\displaystyle \frac{actualvapordensity}{saturationvapordensity}}\times 100\mathrm{\%}$ 
 Step 2 : Here we are given the following datas , 
              actual vapor density = 10g/m3$10g/m^3$
              saturation vapor density = 17.3g/m3$17.3g/m^3$
              Putting these in the given formula, Relative Humidity=1017.3×100%$RelativeHumidity={\displaystyle \frac{10}{17.3}}\times 100\mathrm{\%}$
                                                                                                              =57.8%$=57.8\mathrm{\%}$
               

Definition

Relative Humidity

It is the ratio of the amount of water (moisture) present in air to the maximum amount of water which can be absorbed by air. 
It is expressed as a percentage.It is a function of both moisture content and temperature.

Result

Effect of pressure on boiling point

The boiling point of a liquid increases with increase in pressure. Hence, cooking at high altitudes is difficult.

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