5.9 and 5.10 answers
Saturday, October 29, 2011
5.7 and 5.8
5.7 and 5.8 
· 5.7 understand that a substance can change state from solid to liquid by the process of melting
· 5.8 understand that a substance can change state from liquid to gas by the process of evaporation or boiling
· Questions from Collins p.112
· Answer in Bullet Points!
Q1)
a)
- The bonds in a solid are very strong and so the particles are held into place. This does not allow any of the particles to freely flow around and they can only vibrate on the spot. Since they are fixed, their shape cannot be altered.
- Liquids and gases can move and slide over each other because their bonds are weaker than a solid. Therefore, they do not keep their shape.
b)
- The particles in a solid and a liquid are closely packed together and therefore, incompressible.
- The particles in a gas are widely spaced and the bonds between them are much more weaker than a solid and liquid, so they can fill their container.
Q3)
Boiling:
- Boiling is when you heat up a liquid until the average kinetic energy is enough to turn it into a gas
- This only occurs at a fixed temperature, the boiling point of the liquid
- Boiling happens throughout liquids and is a fast process.
Evaporating:
- Happens when a liquid is left open in the air
- Only the particles at the surface of the liquid have the ability to escape from the liquid and into the air
- Evaporation occurs at a range of temperatures. A high temperature would increase the rate of evaporation whereas a low temperature would decrease the rate of evaporation.
- Because most of the energetic particles have been removed, the average kinetic energy decreases and the liquid cools down.
5.7 and 5.8 Experiment - Cooling Curve of Stearic Acid using datalogger
5.7 to 5.10 Plenary 1
· Play the Stage 1 game to test your knowledge of solids, liquids and gases
· Play the Stage 2 game to test your knowledge about changes of phase!
5.7 to 5.10 Plenary 2
Play the Level 1 game to test your knowledge of the properties of solids, liquids and gases
Extension: Play the Level 2 game to extend your knowledge about changes of phase!
Saturday, October 22, 2011
5.6 Questions
Note: ρfresh water = 1,000kg/m3; g = 10N/kg
Q5) ∆Pressure = height x density x gravitational field strength
250,000 - 100,000 = h x 1,000 x 10
h = 15m
If he were diving in sea water that is slightly denser than fresh water, the height would increase.
Q6) ∆Pressure = height x density x gravitational field strength
∆Pressure = 50 x 420 x 1.4
∆Pressure = 29.4kPa
1600 mbar = 160 kPa
160+29.4 = 189 (189.4) kPa
Wednesday, October 12, 2011
Topic 5 - 5.6 Demo - squirting water column
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· The bottom hole squirts water the furthest
· Because the water at the bottom has the greatest pressure
· Because in the formula ∆p = h × ρ × g, ρ is constant, g is constant and h is large
· So ∆p = large
Topic 5 - 5.6
· 5.6 recall and use the relationship for pressure difference:
pressure difference = height × density × g
∆p = h × ρ × g
∆p = pressure of the fluid (N/m2 or Pa)
h = height of the fluid (m)
ρ = density of the fluid (kg/m3)
g = gravitational field strength (N/kg)
Topic 5 - 5.5 Demo 1 - Magdeburg Hemispheres
· Magdeburg Hemispheres
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· And here are the horses I was talking about! http://www.youtube.com/watch?v=7bJkaFByiA0&feature=related
Topic 5 - 5.5
· 5.5 understand that the pressure at a point in a gas or liquid which is at rest acts equally in all directions
Topic 5 - 5.4 Starter 2 explained
5.4 Starter 2 explained·Your finger pushes on the pin and the pin pushes back on your finger· N3L tells us that all these two forces are equal in size· The pin pushes on the wall and the wall pushes back on the pin· N3L tells us that all these two forces are also equal in size· If the surface area is large then the force is spread over a large area and the pressure is low· If the surface area is small then the force is spread over a small area and the pressure is high· You would like the pressure on your finger to be low and the pressure on the wall to be high· The other way round is painful!
Topic 5 - 5.4 Harder questions on Pressure
·Collins, p.107, Q.4.
Q4)
Ordinary Shoe heel:
Area: 0.0025m^2
Mass: 40x10 = 400N
Pressure = Force / Area
Pressure = 400 / 0.0025
Pressure = 160,000 N/m^2
- - - - - - - - - - - - - - - - - - - - - - -
Elephant:
Area = πx0.1x0.1
Mass = 40x10 = 400N
Pressure = Force / Area
Pressure = 5000 / ( πx0.1x0.1)
Pressure = 160,000 N/m^2 (2 s.f.)
- - - - - - - - - - - - - - - - - - - - - - -
High-heeled shoe:
Area = 0.5/10,000
Mass = 400
Pressure = Force / Area
Pressure = 400 / (0.5/10000)
Pressure = 8,000,000 N/m^2
- - - - - - - - - - - - - - - - - - - - - - -
The high-heeled shoe will damage a wooden floor that starts to yield at a pressure of 4000 kPa (4000 x 1000 = 4,000,000 N/m^2).
Tuesday, October 11, 2011
Topic 5 - 5.4
·5.4 recall and use the relationship between pressure, force and area:pressure = force / area
p = F / A
Thursday, October 6, 2011
Topic 5 - 5.2 Harder Questions
Q1)·Collins p.106 Q.1-3. (Table of densities below)
Wood in oil = It depends on what sample of the material you choose.
Wood in mercury = Float
Plastic in oil = Sink
Steel in mercury = Float
Silver in air = Sink
Gold in mercury = Sink
Helium balloon in air = Float
Q2)
Brick dimensions = 0.2m x 0.09m x 0.065m
Weight of brick = 2.22kg
Density = mass / volume = 2.22 / 0.00117 = 1,897kg/m^3
Q3)
If the crown is pure gold, the new water level would be 800 + 100 = 900 cm^3. This will be because the volume of the gold would be m / Density, V = 1930 / 1.93, V = 100 cm^3. The water level would not be as much because silver is less dense than gold. Therefore, the volume would be less.
Topic 5 - 5.3 Plenary 2
How can you make a cannonball float?
Explanation below…Density of Iron = 7.9g/cm3 or 7,900kg/m3Density of Mercury = 13.6g/cm3 or 13,600kg/m3
Topic 5 - Density of an irregular solid
Results:
1. m = 109 g
2. V = 36 cm^3
3. ρ = m/V = 109 / 36 = 3.03 g/cm^3
1. m = 109 g
2. V = 36 cm^3
3. ρ = m/V = 109 / 36 = 3.03 g/cm^3
Topic 5 - Density of a liquid
Results:
1. m(cylinder) = 50g
2. m (cylinder+liquid) = 68g
3. m(liquid) = m(cylinder+liquid) - m(cylinder) = 68 - 50 = 18g
4. V = 19.5 ml = 19.5 cm^3
5. Density = m/V = 18/19.5 = 0.92 g/cm^3
1. m(cylinder) = 50g
2. m (cylinder+liquid) = 68g
3. m(liquid) = m(cylinder+liquid) - m(cylinder) = 68 - 50 = 18g
4. V = 19.5 ml = 19.5 cm^3
5. Density = m/V = 18/19.5 = 0.92 g/cm^3
Wednesday, October 5, 2011
Topic 5 - 5.2 Plenary Answer
5.2 Plenary Answer04 October 201114:16·If you take a piece of wood that's got a density of 2.4g/cm3 and cut it exactly in half, what will the density of each of the 2 new pieces of wood be?
· 2.4g/cm3!
·You've halved the mass of each block but you've also halved the volume of each block, so the ratio m/V remains contant!· For example…· ρbig block = m/V = 240/100 = 2.4g/cm3· ρsmall block = m/V = 120/50 = 2.4g/cm3
Topic 5 - 5.2
·5.2 recall and use the relationship between density, mass and volume:density = mass / volume
ρ = m / V
Topic 5 - Obj. 5.3
Density of a regular solid
Method:
1. Measure mass using a balance
2. Measure width, length and height using a ruler
3. Calculate volume (V = w x l x h)
4. Calculate density (ρ = m / V)
Method:
1. Measure mass using a balance
2. Measure width, length and height using a ruler
3. Calculate volume (V = w x l x h)
4. Calculate density (ρ = m / V)
Results:
1. m = 580.4 g
2. w = 5 cm; l = 5 cm; h = 2 cm
3. V = w x k x g = 5 x 5 x 2 = 50cm^3
4. ρ = m / V = 580.4 / 50 = 11.608 g/cm^3
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