PDF- Calorimetery, COF ENOPLUG ELECTROMAGNETIC CALORIMETER -Non-isothermal kinetic study of the thermal decomposition of N-{bis - CALORIMETERY

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## Mohammed Asif

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CALORIMETERY → The questions asked on calorimetery involves mainly those questions in which phase change of water is involved

You are given steam,

ice and water mixture and you are supposed to calculate the final temperature and composition of the mixture

We are dealing all possible questions here along with their magical shortcut solutions

SOME BASIC – DEFINATIONS

SPECIFIC HEAT (OR) SPECIFIC HEAT CAPACITY:

It is defined as the quantity of heat required to raise the temperature of 1 kg of substance by 10 C

• dQ S= Sp

Heat is found to maximum for HYDROGEN 3

• 5 and then for water M

dT cal 1 gm − k Large Sp

Heat of water means it will take longer time to heat up water and water will take longer time to cool drum

Coastal areas happen to have temperature climate (that means not much variations in day and night temperature) because of land breeze and see breeze

### And reason for land and sea breeze is large Sp

• heat of water and low Sp

### Wt and the Sp

Heat of an element at constant volume is constant for a nor of elements in solid state and is equal to constant value 5

• 96 cal mole− k

#### It means heavies the elements,

• lesser will be the Sp

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• ∴ H=MS

## It means Let us suppose we have to pieces of copper of diff

mass then both will have same Sp

Heat but different heat capacity

C is given by reciprocal of slope of heat temp curve

1 ∴ H

### C at point P = tan θ

• → MOLAR – HEAT CAPACITY:
• - It is the heat required to raise the temperature of one mole ofa substance by 10 C (or) 1 k

J Unit :

• - mole− k

### E Water equivalent = Heat Capacity →

LATENT HEAT (OR) HEAT TRANSFORMATION: Whenever a body is under going its phase change its keeps absorbing heat without change in its temperature

So the amount of heat required to change the state of unit mass of a substance at a const temp is called the heat of transformation 9or) Latent heat (L)

#### Q=ML ∴

Latent heat of friction L'f

For Ice L'f = 80cal / gm Latent Heat

for water Latent heat of vaporization ( L'v ) L'v = 540cal / gm

REAL – LIFE – EXAMPLE 1) If we have to extinguish a fire we should throw boiling water on it instead of cold water because the boiling water has the tendency to get converted into steam during which one gram water can absorbs 540 cal / gm

But if say we have cold water at 00 C then it takes a very long time to heat it up and that too upto 1000 C it will absorbed only 100 cal per gm

Now that does not mean if you have a burn in your finger then you’ll put your finger in boiling water

You should place your finger in the coldest water because here the heat of your finger is being released through the process of conduction and larger the temperature difference between your finger and outside cold medium longer will be heat transfer

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• 2) When we say that an air – conditioner is of 1
• 5 ton capacity it means that if that AC runs for 24 hrs,

continuously it will convert 1

• 5ton water at 00 C

So 1 ton AC has a power of approximately 3

9 kilowatt

## Remember “ton” of AC is the commonly used unit of power of AC and not its weight

• 3) Conversion of vapour directly into solid is known as “HOAR – FROST” Eg:
• - formation of snow by freezing of clouds in low temp zone

# So Hoar frost is a phenomenon reverse of sublimation

• 4) Melting of ice under pressure and its be coming solid again when the pressure is released is called “REGELATION”

PRINICIPLE OF CALORIMETRY:

#### When a no

of bodies are mixed together in a container in which there is no heat loss to the surroundings,

the heat lost by a group of bodies is equal to the heat lost by rest of the bodies

The final temperature of the mixture can never be greater then the highest temperature of the components and it can’t be lower than the lowest temp of the components

When a body doesn’t undergo its phase change use the equation Q = ms ∆t ,

and when it undergoes the phase change use the equation Q = ML

When we say that heat lost by one body = heat gained by other body,

it doesn’t necessarily mean that temperature lost by one body is equal to the temperature gained by the other body

Question:

• - A solid material is supplied with heat at a constant rate

#### The temp of the material is changing with the heat input as shown

• 1) What do the horizontal regions AB and CD represents
• 2) If CD = 2AB,
• what do you infer

? 3) What does the slope of DE represent 4) The slope of OA > the slope of BC

What does this indicate

# C T e m

• i n p u t

## Solution: a) AB = M

• p CD = B

p b) CD = 2 AB means latent heat of vaporization is twice that of latent heat of M

Tempchange c) slope of DE is basically which is basically inverse of heat Heat input capacity so DE is inverse of heat capacity of gaseous phase

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# BC is inverse of heat capacity of liquid phase

## OA is inverse of heat capacity of solid phase

d) Since slope is inversely proportional to heat capacity and DA has larger slope

So OA portion has less heat capacity (or) you can say that the solid sate has less heat capacity compared to the liquid state

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# Q) 5 gm ice at oo C is mixed with 1 gm steam at 100 0 C

#### Find the final temp and composition of the mixture

• - In such questions assume that finally we get all water at t0 C

This phenomenon can be represented diagrammatically

Here ice will gain heat and steam will rebase it

• 5 g m IC E 1 g m s't Heat gained = Heat lost 0 0 0C 1 0 C0 ML f + MS ∆t = ML v + MS ∆t
• c o n s'e n

p s'a t io n 5 X 80 + 5 X 1 X (t – 0) = 1 X 540 + 1 X 1 X (100 – t) h e a t i n c'go o l'i n g 400 + 5 t = 640 – t w a t e 1 r 0 0 C0 00C t0c 6 t = 240 p h a s'e T = 400 C Since all 5 gm ice and 1 gm steam has become water so finally we will get 6 gm water at 400 C

## Q) 100 gm ice at 00 C is mixed with 10 gm steam at 100 0 C

### Find the final temperature and composition

• - Heat lost by stem = Heat gained by ice Heat lost = heat lost by steam in condensation + heat cost in cooling form 1000 C water to t0 C

∴ Heat lost = 10 X 540 + 10 X 1 X (100 – t) = 5400 + 1000 – 10 t = (6400 – 10 t) cal Heat gained = Heat gained by ice in melting + heat gained by water form 00 C to t0 C

Heat gained = 100 X 80 + 100 X 1 X (t – 0) = (8000 + 100 t) cal ∴ (6400 – 10 t) = 8000 + 100 t 1600 t=− = − 14

• 540 C 110 This – ve temp value is an absurd answer because if two substances mixed have temp 00 and 1000 C then final temp will be either 00 or 1000 C (or) in between

### How can it be less than 00 C

! So here – ve temp indicates that complete ice is not melt and we will get ice – water mixture at 0 0 C

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• ∴ Amount of icemelt =

### Heat lost by 10 gm steam in reaching 00 C water

L ice 10× 540+ 10× 1× ( 100 − 0) = = 80gm 80 So ice left = (100 gm – 80 gm) = 20 gm Total water= 80 gm from ice + 10 gm form steam = 90 gm

#### Q) 30 gm ice at 00 C is mixed with 25 gm steam at 100 0 C

Find the final temperature and composition

• - Heat lost by steam upto t0 C water = Heat gained by ice a upto to C water 25 X 540 + 25 X (100 – t) = 30 X 80 + 30 X 1 X t t = 2630 C The final temperature being more than 1000 C is an absurb value because if two substan less at 00 C and 1000 C are mixed then the final temp should be 0 0 C or 1000 C (or) in between

Here the final answer more than 1000 C indicator that the steam is not condensed completely rather we have water steam mixture at 1000 C

# The amount of steam condensed = =

Heat gained by iceupto 1000 Cwater

• 30× 80+ 30× 1× 100
• = 10 gm

L steam

So steam left = (25 – 10) = 15 gm Amount of water = 30 gm from ice + 10 gm from steam = 40 gm So finally we have 40 gm water and 15 gm stem at 1000 C

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