2ND TERM
Posted: Wed Jun 24, 2015 12:27 pm
Sub –topic 2: 1. Experiment to determine the linear expansivity of a metal block.
2. Area expansivity
1. Experiment to determine the linear expansivity of a metal block.
Aim: Experiment to determine the linear expansivity of a metal block
Apparatus: Thermometer, Micrometer screw guage, steam jacket, metal rod, meter rule
Method: i. Measure the length of the metal rod (L1).
ii. Insert the metal rod in the steam jacket and take the initial temperature of the
metal rod with thermometer (θ1).
iii. Screw the micro-meter to touch the end of the rod and take the reading of
micro-meter (xi).
iv. Unscrew micro meter to make room for expansion of metal rod.
v. Introduce steam into the steam jacket for several minutes then the metal rod
will expand.
vi. Screw the micrometer screw guage to touch the end of the metal rod again and take
the reading again (x2).
vii. Record the final temperature (θ2).
Calculation:
α = (¬x_1- x_2)/(L_(1 ( θ_2- θ_(1)) ) )
Conclusion: Since all parameters are known α can be calculated.
2. Area or Superficial expansivity(β)
It is defined as the increase in area per unit area per degree rise in temperature
β = (Increase in Area)/(original Area x temperature rise ) = (¬A_2- A_1)/(A_(1 (θ_2- θ_1 )_ ) )
A_2- A_1=increase in Area or expansion
θ_2- θ_(1=Temperature rise or increase in temperature)
θ_1 is initial temperature, θ_2 is final temperature
A_1=original Area, A2 is new area
Relationship between Linear expansivity and Area expansivity:
Β = 2α
Question 1: A metal cube of cross sectional area 3.45m2 at 00c is heated at a temperature rise of 70k, when the final length of the cube is 3m. Find the: i. coefficient of superficial expansivity.
ii. coefficient of linear expansivity.
Solution:
β = (Increase in Area)/(original Area x temperature rise ) = (¬A_2- A_1)/(A_(1 (θ_2- θ_1 )_ ) )
A_2= L^(2 )= 3 x 3 = 9m2
θ_2- θ_1=70K
A_1=3.45m2
β = (¬9-3.45)/( 3.45 x 〖70〗^ )= 5.55/241.5=0.023/k=2.3 x 〖10〗^(-2)/k
EVALUATION:
1. The linear expansivity of a metal is 0.000019 per k. What will the area of 400mm2 be if its temperature is raised by 100C.
Sub –topic 3: Cubic expansivity, Experiment to determine the apparent cubic expansivity.
Cubic or volume expansivity(γ)
It is defined as the increase in volume per unit volume per degree rise in temperature
γ = (Increase in volume)/(original volume x temperature rise ) = (¬V_2- V_1)/(V_(1 (θ_2- θ_1 )_ ) )
V_2- V_1=increase in volume or expansion
θ_2- θ_(1=Temperature rise or increase in temperature)
V_1=original volume V_2 is final volume.
Relationship between Linear expansivity and Cubic expansivity:
γ = 3α
Question 2: The increase in the volume of 10cm3 of mercury when the temperature rises by 1000c is 0.182cm3. What is cubic expansivity of mercury.
γ = (Increase in volume)/(original volume x temperature rise ) = (¬V_2- V_1)/(V_(1 (θ_2- θ_1 )_ ) )
V_2- V_1=increase in volume=0.182cm2
θ_2- θ_(1=Temperature rise=) 〖100〗^o
V_1=original volume=10cm2
γ = (¬0.182)/( 10 x 〖100〗^ )= 0.182/1000=0.000182/k=1.82 x 〖10〗^(-4)/k
EXPANSION IN LIQUIDS:
Expansion in liquid is complicated by the expansion of the container because while the liquid expands the container equally expands. So it is important to differentiate between real and apparent cubic expansivity.
REAL OR ABSOLUTE CUBIC EXPANSIVITY (γr): It is defined as the increase in volume per unit volume per degree rise in temperature.
APPARENT CUBIC EXPANSIVITY (γa): It is defined as the increase in volume per unit volume per degree rise in temperature when the liquid is heated in an expansible vessel.
γ_r=γ_a+ γ
γ_r=Real or cubic expansivity of liquid
γ_a=Apparent cubic expansivity of liquid
γ=cubic expansivity of the container
QUESTION 3:
A cube with side 100cm at 00C is heated to 1000C. If the side becomes 101cm long find,
a. The linear expansivity
b. The cubic expansivity
SOLUTION:
〖(a). L〗_1= 100cm= 1m, L_2=101cm=1.01m, θ_2=〖100〗^0,θ_1=0^0
α = (Increase in lenght)/(original lenght x temperature rise ) = (¬L_2- L_1)/(L_(1 ( θ_2- θ_(1)) ) )
α = (¬L_2- L_1)/(L_(1 ( θ_2- θ_(1)) ) )
α = (¬1.01-1)/( 1x 〖100〗^0 )= 0.01/100=0.0001=1.0 x 〖10〗^(-4)/k
(b). γ = 3α
γ = 3 x 1.0 x 10-4= 3.0 x 10-4/K
DETERMINATION OF THE APPARENT CUBIC EXPANSIVITY OF A LIQUID.
APPARATUS: Thermometer, Density bottle, Retort stand, Water, Source of heat, Beaker, Beam balance, Liquid, Stirrer.
METHOD:
i. Dry the density bottle and weigh it (M).
ii. Fill the density bottle with the liquid that the apparent cubic expansivity is
required and weigh it (M1)
iii. Immerse the density bottle into a beaker of water and suspend with a thread on
the clamp of the retort stand.
iv. Take the original temperature of the water in the beaker (θ1).
v. Heat the set up gently until the water boils.
vi. Some liquid are expelled through the orifice of the bottle cover, the heating
continues until no liquid is seen expelled again.
vii. The final temperature of water is taken(θ2)
viii. The density bottle is removed and wiped dry and re-weighed(M2).
CALCULATION:
Mass of empty density bottle = M
Mass of density bottle + liquid = M1
Original temperature of water = θ1
Final temperature of liquid = θ2
Mass of remaining liquid + density bottle = M2
γ_a = █(@Mass of liquid expelled)/(Mass of liquid remaining x temperature rise ) = (¬M_1- M_2)/(M_(2-M (θ_2- θ_1 )_ ) )
CONCLUSION:
Since all the parameters are known, apparent cubic expansivity γ_a can be
calculated.
EVALUATION:
1. Differentiate between Real cubic expansivity and Apparent cubic expansivity.
2. A glass bottle full of mercury has mass 500g on being heated through 350C, 2.43g of mercury was expelled. Calculate the mass of mercury remaining in the bottle ( cubic expansivity of mercury is 1.8 x 10-4/k and linear expansivity of glass is 8.0 x 10-6/k)
Sub-Topic 4: Applications of Expansion
(A) ADVANTAGES OF EXPANSION:
1. The use of the bimetallic strip in:
a. Fire alarm
b. Bi-metallic thermometer
c. Electric pressing iron
2. Red hot rivet used in ship
3. Removal of tight glass stopper
4. Fitting of wheels in rims
Bi-metallic strip:
It consists of two different metals joined together. They expand at different rates when heated e.g brass and iron.
a. ELECTRIC FIRE ALARM:
When a fire breaks out in a building, the resulting heat causes the bi-metallic strip to bend towards the contact, thus completing the circuit. This causes the bell to ring out a fire alarm.
b. BI-METALLIC THERMOMETER
It consists of a coiled bi-metallic strip which expands outwards when heated. As this
happens, the pointer moves along the scale and the reading on the scale is taken as the
temperature.
c. ELECTRIC PRESSING IRON:
It has a device known as Thermostat, it is made of bi-metallic strip and it is used to regulate
the temperature of the pressing iron, Gas cooker, Refrigerator.
MODE OF OPERATION OF PRESSING IRON
When the current is switched on, current flows through the circuit and the bi-metallic strip
expands and bends away from the contact point thereby switching off the flow of current.
The pressing iron cools down and contact is re-made and current flows again and the process
continues. This make-and-break device regulates the temperature of a pressing iron.
2. RED-HOT RIVET USED IN SHIP:
Steel plates and girders which are used in ship building and other constructional works are usually riveted together.
3. FITTING OF WHEELS IN RIMS:
The large driving wheels of locomotive are fixed with steel tyre which are renewed from time to time as they wear out. In order to ensure a tight fitting, the tyre is made slightly smaller in diameter than the wheel. The tyre contracts on cooling thus ensuring tight fitting.
4. REMOVAL OF TIGHT GLASS STOPPER:
A tight glass stopper can be removed by standing the bottle in hot water. The glass bottle expands and the stopper becomes loose.
Experiment to show that water is a poor conductor of heat.
Aim: To show that water is a bad conductor of heat.
Apparatus: water, test tube, ice-block, Bunsen burner and wire guaze.
Method: i. Wrap the ice block with wire guaze to prevent the ice from floating in water,
and drop in the water in the test tube.
ii. Heat the water near the top of the water with the Bunsen burner.
Observation: It is observed that while the water was boiling on top, the ice at the bottom
did not melt
Conclusion: The ice did because water is poor conductor of heat and was not able to conduct the heat to the ice.
Applications of conductors and insulators
1. Cooking utensils: Bad conductors of heat are used as handles while the cooking pots are made of metals which are good conductors of heat.
2. Lagging: Insulators are often used as lagging materials in hot water pipes, stem boilers, hot water storage tanks and ovens to prevent them from freezing or getting colder.
3. Warmth: Woollen sweaters keep us warm during winter or cold weather to prevent conduction of heat from the body.
4. Double walls: Houses built with double walls with space in between them have air trapped in the spaces that act as insulators, thus, keeping the house warm.
Sub-topic 2: Ways of producing charges.
1. Electrostatic induction
Electrostatic induction is the act of charging a neutral body by placing a charged body near it without any contact between the two.
STEP 1 STEP 2
STEP 3 STEP 4
STEP 1: A negatively charged body is brought near the uncharged body, free electrons from the metal sphere are repelled by the excess electrons on the rod. They shift towards the right. They can not escape from the sphere because the stand and the surrounding air are insulated.
STEP2: These excess charges called induced charges are released to the earth by touching the right part of the sphere with a wire and the other part of the wire to the earth.
STEP 3: The wire is disconnected.
STEP 4: The negatively charged rod is removed. A net positive charge is left on the rod.
2. Area expansivity
1. Experiment to determine the linear expansivity of a metal block.
Aim: Experiment to determine the linear expansivity of a metal block
Apparatus: Thermometer, Micrometer screw guage, steam jacket, metal rod, meter rule
Method: i. Measure the length of the metal rod (L1).
ii. Insert the metal rod in the steam jacket and take the initial temperature of the
metal rod with thermometer (θ1).
iii. Screw the micro-meter to touch the end of the rod and take the reading of
micro-meter (xi).
iv. Unscrew micro meter to make room for expansion of metal rod.
v. Introduce steam into the steam jacket for several minutes then the metal rod
will expand.
vi. Screw the micrometer screw guage to touch the end of the metal rod again and take
the reading again (x2).
vii. Record the final temperature (θ2).
Calculation:
α = (¬x_1- x_2)/(L_(1 ( θ_2- θ_(1)) ) )
Conclusion: Since all parameters are known α can be calculated.
2. Area or Superficial expansivity(β)
It is defined as the increase in area per unit area per degree rise in temperature
β = (Increase in Area)/(original Area x temperature rise ) = (¬A_2- A_1)/(A_(1 (θ_2- θ_1 )_ ) )
A_2- A_1=increase in Area or expansion
θ_2- θ_(1=Temperature rise or increase in temperature)
θ_1 is initial temperature, θ_2 is final temperature
A_1=original Area, A2 is new area
Relationship between Linear expansivity and Area expansivity:
Β = 2α
Question 1: A metal cube of cross sectional area 3.45m2 at 00c is heated at a temperature rise of 70k, when the final length of the cube is 3m. Find the: i. coefficient of superficial expansivity.
ii. coefficient of linear expansivity.
Solution:
β = (Increase in Area)/(original Area x temperature rise ) = (¬A_2- A_1)/(A_(1 (θ_2- θ_1 )_ ) )
A_2= L^(2 )= 3 x 3 = 9m2
θ_2- θ_1=70K
A_1=3.45m2
β = (¬9-3.45)/( 3.45 x 〖70〗^ )= 5.55/241.5=0.023/k=2.3 x 〖10〗^(-2)/k
EVALUATION:
1. The linear expansivity of a metal is 0.000019 per k. What will the area of 400mm2 be if its temperature is raised by 100C.
Sub –topic 3: Cubic expansivity, Experiment to determine the apparent cubic expansivity.
Cubic or volume expansivity(γ)
It is defined as the increase in volume per unit volume per degree rise in temperature
γ = (Increase in volume)/(original volume x temperature rise ) = (¬V_2- V_1)/(V_(1 (θ_2- θ_1 )_ ) )
V_2- V_1=increase in volume or expansion
θ_2- θ_(1=Temperature rise or increase in temperature)
V_1=original volume V_2 is final volume.
Relationship between Linear expansivity and Cubic expansivity:
γ = 3α
Question 2: The increase in the volume of 10cm3 of mercury when the temperature rises by 1000c is 0.182cm3. What is cubic expansivity of mercury.
γ = (Increase in volume)/(original volume x temperature rise ) = (¬V_2- V_1)/(V_(1 (θ_2- θ_1 )_ ) )
V_2- V_1=increase in volume=0.182cm2
θ_2- θ_(1=Temperature rise=) 〖100〗^o
V_1=original volume=10cm2
γ = (¬0.182)/( 10 x 〖100〗^ )= 0.182/1000=0.000182/k=1.82 x 〖10〗^(-4)/k
EXPANSION IN LIQUIDS:
Expansion in liquid is complicated by the expansion of the container because while the liquid expands the container equally expands. So it is important to differentiate between real and apparent cubic expansivity.
REAL OR ABSOLUTE CUBIC EXPANSIVITY (γr): It is defined as the increase in volume per unit volume per degree rise in temperature.
APPARENT CUBIC EXPANSIVITY (γa): It is defined as the increase in volume per unit volume per degree rise in temperature when the liquid is heated in an expansible vessel.
γ_r=γ_a+ γ
γ_r=Real or cubic expansivity of liquid
γ_a=Apparent cubic expansivity of liquid
γ=cubic expansivity of the container
QUESTION 3:
A cube with side 100cm at 00C is heated to 1000C. If the side becomes 101cm long find,
a. The linear expansivity
b. The cubic expansivity
SOLUTION:
〖(a). L〗_1= 100cm= 1m, L_2=101cm=1.01m, θ_2=〖100〗^0,θ_1=0^0
α = (Increase in lenght)/(original lenght x temperature rise ) = (¬L_2- L_1)/(L_(1 ( θ_2- θ_(1)) ) )
α = (¬L_2- L_1)/(L_(1 ( θ_2- θ_(1)) ) )
α = (¬1.01-1)/( 1x 〖100〗^0 )= 0.01/100=0.0001=1.0 x 〖10〗^(-4)/k
(b). γ = 3α
γ = 3 x 1.0 x 10-4= 3.0 x 10-4/K
DETERMINATION OF THE APPARENT CUBIC EXPANSIVITY OF A LIQUID.
APPARATUS: Thermometer, Density bottle, Retort stand, Water, Source of heat, Beaker, Beam balance, Liquid, Stirrer.
METHOD:
i. Dry the density bottle and weigh it (M).
ii. Fill the density bottle with the liquid that the apparent cubic expansivity is
required and weigh it (M1)
iii. Immerse the density bottle into a beaker of water and suspend with a thread on
the clamp of the retort stand.
iv. Take the original temperature of the water in the beaker (θ1).
v. Heat the set up gently until the water boils.
vi. Some liquid are expelled through the orifice of the bottle cover, the heating
continues until no liquid is seen expelled again.
vii. The final temperature of water is taken(θ2)
viii. The density bottle is removed and wiped dry and re-weighed(M2).
CALCULATION:
Mass of empty density bottle = M
Mass of density bottle + liquid = M1
Original temperature of water = θ1
Final temperature of liquid = θ2
Mass of remaining liquid + density bottle = M2
γ_a = █(@Mass of liquid expelled)/(Mass of liquid remaining x temperature rise ) = (¬M_1- M_2)/(M_(2-M (θ_2- θ_1 )_ ) )
CONCLUSION:
Since all the parameters are known, apparent cubic expansivity γ_a can be
calculated.
EVALUATION:
1. Differentiate between Real cubic expansivity and Apparent cubic expansivity.
2. A glass bottle full of mercury has mass 500g on being heated through 350C, 2.43g of mercury was expelled. Calculate the mass of mercury remaining in the bottle ( cubic expansivity of mercury is 1.8 x 10-4/k and linear expansivity of glass is 8.0 x 10-6/k)
Sub-Topic 4: Applications of Expansion
(A) ADVANTAGES OF EXPANSION:
1. The use of the bimetallic strip in:
a. Fire alarm
b. Bi-metallic thermometer
c. Electric pressing iron
2. Red hot rivet used in ship
3. Removal of tight glass stopper
4. Fitting of wheels in rims
Bi-metallic strip:
It consists of two different metals joined together. They expand at different rates when heated e.g brass and iron.
a. ELECTRIC FIRE ALARM:
When a fire breaks out in a building, the resulting heat causes the bi-metallic strip to bend towards the contact, thus completing the circuit. This causes the bell to ring out a fire alarm.
b. BI-METALLIC THERMOMETER
It consists of a coiled bi-metallic strip which expands outwards when heated. As this
happens, the pointer moves along the scale and the reading on the scale is taken as the
temperature.
c. ELECTRIC PRESSING IRON:
It has a device known as Thermostat, it is made of bi-metallic strip and it is used to regulate
the temperature of the pressing iron, Gas cooker, Refrigerator.
MODE OF OPERATION OF PRESSING IRON
When the current is switched on, current flows through the circuit and the bi-metallic strip
expands and bends away from the contact point thereby switching off the flow of current.
The pressing iron cools down and contact is re-made and current flows again and the process
continues. This make-and-break device regulates the temperature of a pressing iron.
2. RED-HOT RIVET USED IN SHIP:
Steel plates and girders which are used in ship building and other constructional works are usually riveted together.
3. FITTING OF WHEELS IN RIMS:
The large driving wheels of locomotive are fixed with steel tyre which are renewed from time to time as they wear out. In order to ensure a tight fitting, the tyre is made slightly smaller in diameter than the wheel. The tyre contracts on cooling thus ensuring tight fitting.
4. REMOVAL OF TIGHT GLASS STOPPER:
A tight glass stopper can be removed by standing the bottle in hot water. The glass bottle expands and the stopper becomes loose.
Experiment to show that water is a poor conductor of heat.
Aim: To show that water is a bad conductor of heat.
Apparatus: water, test tube, ice-block, Bunsen burner and wire guaze.
Method: i. Wrap the ice block with wire guaze to prevent the ice from floating in water,
and drop in the water in the test tube.
ii. Heat the water near the top of the water with the Bunsen burner.
Observation: It is observed that while the water was boiling on top, the ice at the bottom
did not melt
Conclusion: The ice did because water is poor conductor of heat and was not able to conduct the heat to the ice.
Applications of conductors and insulators
1. Cooking utensils: Bad conductors of heat are used as handles while the cooking pots are made of metals which are good conductors of heat.
2. Lagging: Insulators are often used as lagging materials in hot water pipes, stem boilers, hot water storage tanks and ovens to prevent them from freezing or getting colder.
3. Warmth: Woollen sweaters keep us warm during winter or cold weather to prevent conduction of heat from the body.
4. Double walls: Houses built with double walls with space in between them have air trapped in the spaces that act as insulators, thus, keeping the house warm.
Sub-topic 2: Ways of producing charges.
1. Electrostatic induction
Electrostatic induction is the act of charging a neutral body by placing a charged body near it without any contact between the two.
STEP 1 STEP 2
STEP 3 STEP 4
STEP 1: A negatively charged body is brought near the uncharged body, free electrons from the metal sphere are repelled by the excess electrons on the rod. They shift towards the right. They can not escape from the sphere because the stand and the surrounding air are insulated.
STEP2: These excess charges called induced charges are released to the earth by touching the right part of the sphere with a wire and the other part of the wire to the earth.
STEP 3: The wire is disconnected.
STEP 4: The negatively charged rod is removed. A net positive charge is left on the rod.