Chapter 11. HEAT 2ND YEAR PHYSICS COURSE SINDH FULL NOTES

Saturday, December 5, 2009

Chapter 11. HEAT 2ND YEAR PHYSICS COURSE SINDH FULL NOTES

Question: Define heat and its unit. Define temperature what the scale used to measure the temperature are.
Answer.
Heat
Heat is a form of energy which flow from hot body to cold body when they are in thermal contact.
Or heat is a form of energy which is related with the average movement of the molecule.
Unit
Heat is measured in Calorie (Calorie is the name of a scientist)
Definition of calorie
The amount of heat required to raise 1⁰C temperature of 1 gram water. In MKS system the unit of heat is joule 1 calorie =4.2 joule
Temperature
The degree of hotness or coldness of a body is called as temperature.
Unit: the unit of temperature is Kelvin it is measured by the help of thermometer.
There are three scale used to measure the temperature
1. Centigrade scale
On this scale the lower fixed point is at 0⁰C and upper fixed point is at 100⁰C where interval between these two points is divided in 100 equal scales. Each scale measures 1⁰C temperature
⁰C= 5/9 (⁰F -32)
2. Fahrenheit scale
On this scale the lower fixed point is at 32⁰F and upper fixed point is at 212 ⁰F where the interval between these two points is divided in 180 equal scales each scale measures 1⁰F temperature.
⁰F = 9/5 (⁰C +32)
3. Kelvin or absolute scale
On this scale the lower fixed point is at 273 k and upper fixed point is at 373 k where the interval between these points is divided in 100 equal scales each scale measure 1k temperature.
K =⁰C +273
Related problem conversation of temperature at different scale
Problem11.1
A normal body temperature is 98.4⁰F Find its temperature at ⁰C? And at K scale?
Sol. Data Fahrenheit temperature Tf = 98.4 ⁰F
Centigrade temperature Tc =?
Kelvin temperature K =?
We know that
Tc = 5/9 (F-32)
Tc = 5/9 (98.4-32)
Tc =5/9 (66.4)
Tc =36.89⁰C …………..1
K = 273 +⁰C
K = 273+ 36.89
K = 309.89⁰F …………..2
Question: What is thermal expansion explain linear expansion? Or prove that L’ = L (1 +αΔT)
Answer: Thermal expansion
If a body expand by gaining some heat from external source its dimension expands more due to the temperature difference the phenomena is called thermal expansion. The volume of body increases due to increasing movement of its molecule.

Linear expansion
The increase in length of a body due to heat is called as linear expansion for example if the length of a rod increases by supplying heat its expansion is linear expansion.
Explanation
Consider a thin rod of uniform cross section area
At temperature T1 ⁰C its length is L
At temperature T 2⁰C its length become L’
The increase in length ΔL of rod is directly proportional to the initial length of the rod L and to the change in temperature ΔT
Therefore
ΔL ∝ L …………..1 ΔL = L’-L change in length
ΔL ∝ Δ T ………………..2 ΔT= T2- T1 change in temperature
Combining 1 and 2

ΔL ∝ LΔT
ΔL= α LΔT ………….3
Where α is a sign of proportionality constant and is called coefficient of linear expansion it unit is 1/K.
The value of coefficient of linear expansion depend upon the nature of material the rate of expansion is different of different material
From eq. 3
ΔL= α LΔT
L’ - L= α LΔT
L’= L + α LΔT
L’ = L (1+ α ΔT) proved

Related problem
Question:
The length of a steel rod is 0.2 cm at 30⁰C what is its final length at 60⁰C?
Solution:
initial length L = 0.2cm
initial temperature T1 =30 ⁰C
final temperature T2 =60⁰C
value of coefficient of linear expansion α for steel = 11✕10 -6 /⁰C
final length L’ = ?
L’ = L (1+ α ΔT) ΔT= T2- T1 change in temperature
L’ = 0.2 ( 1+ 11✕10 -6✕ (60-30))
L’ = 0.2 ( 1+ 11✕10 -6✕30)
L’ = 0.2 ( 1+330 ✕10 -6)
L’ = 0.2 ✕1.000330
L’ =0.200066 cm the final length of the rod


Question: What is volume expansion prove that V’ = V (1 +βΔT)
Answer.
When the volume of a body increases by gaining heat it is called as volume expansion the body`s length, width and height increases.
Explanation
Consider a body of volume =V
At temperature =T1⁰C
Increasing temperature from T1⁰C
To T2⁰C at temperature T2⁰C volume become V’
The change in volume ΔV of the body is directly proportional to its initial volume V and change in temperature ΔT

ΔV ∝ V ……..1 → ΔV = V’ - V
ΔV ∝ΔT …………2 → ΔT = T2 –T1
Combining 1& 2
ΔV = constant vΔT
ΔV = β vΔT ……..3
where β is the constant of proportionality and is called coefficient of volume expansion its unit is 1/⁰C or 1/K
Coefficient of volume expansion depends upon the nature of material different materials expand at different rate and have different value of β
From 3 we know that
ΔV = β vΔT
V’ - V = β vΔT
V’ = V + β vΔT
V’ = V (1 +β ΔT) proved
Question: What is the Relationship between linear and volume expansion? or
Prove that β = 3α
To find the relation between α and β consider a box of rectangular shape. h
At temperature T1⁰ C its length is = l
Its width = b b
Its height = h l
The volume of the body at T1⁰ C V= l ✕ b ✕ h ……… A
When we increase its temperature from T1⁰ C to T2⁰ C thermal expansion takes place in the body it expands three dimensionally and its length, width and height increases so that its volume become larger
The length increases as l’ = l (1+ α ΔT) ……1 l’ is the final length
Width increases as b’ = b (1+ α ΔT) ……2 b’ is the final width
Height increases as h’=h (1+ α ΔT) …….3 h’ is the final height
Whereas the final volume at temperature T2⁰ C V’= l’ ✕ h’✕ b’ ….4
Putting the value of l’ ✕ h’✕ b’ from 1, 2 and 3 in 4
V’ =l (1+ α ΔT) ✕ h (1+ α ΔT) ✕ b (1+ α ΔT)
V’= l ✕ h✕ b (1+ α ΔT)3 but from A we know that V = l ✕ h✕ b
V’= V (1+ α ΔT)3 …………B
But the final volume expansion of the body(as the coefficient of volume expansion)
V’= V (1+ β ΔT) …………C
Comparing B and C we get
V (1+ β ΔT)= V (1+ α ΔT)3 canceling V from both sides
(1+ α ΔT)3 = 3 α ΔT + 3 α2 ΔT2+ α3 ΔT3
Neglecting the values of α3, α2 because these are the smallest values
(1+ α ΔT)3 = 1+ 3α ΔT
From C we know that (1+ α ΔT)3 = (1+ β ΔT)
(1+ β ΔT)= 1+ 3α ΔT
1- 1+ β ΔT-3α ΔT=0
ΔT (β -3α)=0
β -3α = 0
β = 3α proved


Related problem
An aluminum sphere of radius 0.4 m is heated from 0 ⁰C to 100 ⁰C what will be the change in its volume?
Solution
Radius of sphere r = 0.4m
Initial temperature T1= 0 ⁰C
Final temperature T2 = 100 ⁰C
Change in volume ΔV=?
Coefficient of linear expansion for aluminum α= 24✕ 10-6/⁰C
β = 3α
Therefore coefficient of volume expansion β will be= 3 ✕24✕ 10-6/⁰C
= 72 ✕ 10-6/⁰ Initial volume of the sphere V1= 4/3 π r3
V1= 4/3 π (0.4)3
V1 =0.268m3
Change in volume of sphere ΔV = β v1 ΔT
ΔV = 0.268✕72✕10-6 (100-0)
ΔV = 1.930 ✕10-3
ΔV = 0.00193 m-3
Therefore the change in volume of aluminum sphere is
0.00193 m-3
Uses of thermal expansion
Bimetallic strip
It is made by joining two different metal strip or alloys in length wise outer side is made of that metal which has higher value of linear expansion. Different type of metals expand at different rate (because the value of coefficient of linear expansion α is different for each metal) therefore when this bimetallic is heated one metal expand more than other. These two metals are joint together firmly therefore bimetallic turns in curve shape. The expansion of bimetallic in curve shape is used in different instrument as for temperature control bimetallic thermostat is commonly used. It is used electric iron aircondtionar, refrigerator, room heater, cars radiator fans, oven and fire alarm.

Heat capacity

The amount of heat require to increase the temperature 1 ⁰ C is called heat capacity

C = ΔQ/ ΔT the value of heat capacity depend upon the nature of material and mass greater the mass of material greater will be the heat capacity J/ ⁰ C

Specific heat capacity

The amount of heat require to increase the temperature 1⁰ C of unit mass of a substance is called specific heat capacity.

More heat energy is required to increase the temperature of a substance with high specific heat capacity than one with low specific heat capacity. For instance, eight times the heat energy is required to increase the temperature of an ingot of magnesium as is required for a lead ingot of the same mass. The specific heat of virtually any substance can be measured, including chemical elements, compounds, alloys, solutions, and composites

equations

• The equation relating heat energy to specific heat capacity, where the unit quantity is in terms of mass is:

where Q is the heat energy put into or taken out of the substance, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

• Where the unit quantity is in terms of moles, the equation relating heat energy to specific heat capacity (also known as molar heat capacity) is

where Q is the heat energy put into or taken out of the substance, n is the number of moles, c is the specific heat capacity, and ΔT is the change in temperature.

C` = ΔQ/ mΔT

It only depends on the nature of the material not on the mass of the material. Each material has different specific heat capacity. The specific heat capacity of water is 4200j/kg.

Gas laws
Boyle`s law
“The volume of a given mass gas is inversely proportional to its pressure at constant temperature”
P ∝ 1/ V when T = constant
PV = constant T = constant
If the pressure of a gas is increased from P1 to P2 at constant pressure it volume will increase from V1 to V2 then according to Boyle`s law
P1V1 = P2V2
Mass effect: at given temperature pressure of gas P and volume of gas V depend upon mass of gas. Therefore
PV ∝ m
PV/m = constant
P1V1 /m1=P2 V2 /m2 = CONSTANT at temperature constant .

Charles`s law

It states that

“The volume of a given mass gas is directly proportional to its absolute temperature at constant pressure”

V ∝ T when P (pressure)= constant

P/V= CONSTANT when P(pressure)= constant

P1v1 =P_2/ T₂ at P= CONSTANT

General gas equation
If at constant temperature T1 the pressure of the gas is changed from P1 to P2 so that its volume is changed from V1 to V than according to Boyle’s law
P1 V1 = P2 V
V = P1 V1/ P2 …….A
Now at constant pressure P2 changing its temperature from T1 to T2
Its pressure changed from V to V2 than according to Charles law
V/T1 = V2/T2 …………B
Putting the value of V from equation A in equation B
We have
P1V1/P2 /T1 = V2/T2

P1V1/T1 = P2 V2/T2
PV/T = Constant


Relation between heat capacity and molar heat capacity
Consider M is the molecular weight of a substance the mass of n mole is m
m = nM
Specific heat capacity of the substance
C` = ΔQ/ mΔT
C` = ΔQ/ nMΔT
MC`= ΔQ/ nΔT C = ΔQ/ nΔT
MC` = C