Assignment #1 - Due Oct. 18th 2007

( there is a ½ mark deduction for each answer that does not have a correct unit to a maximum of 5 marks)

  1. Chapter 2 Question 2 (discuss means tell me what transformations are happen) (page 57) Problem 3, 15 page 59-60, Problem 12 page 74

Question 2

Discuss the energy transformations that are happening in the following devices or events. (7 marks)

  1. striking a match - potential chemical energy to light and heat energy (1 mark)
  2. windmill- kinetic energy of the wind to mechanical energy to turn the windmill (or to electrical energy if attached to a turbine) 1 mark
  3. ball rolling off a table and bouncing until it stops – this is a series of potential (as the ball goes down) to kinetic and then kinetic a to potential (as the ball goes up (transformations) (2 marks). The ball stops because it is constantly losing energy to friction (1 mark)
  4. microphone - sound energy is being converted to electrical energy (1 mark)
  5. flashlight – chemical energy (from the battery) is being converted to light (1 mark).

Problem 3 (2 marks)

What is the potential energy of a 70 kg person sitting on a ladder 2 m off the ground?

m = 70 kg

g=9.8

h =2 m

PE = mgh

= 70 x 9.8 x 2 (1 mark for showing calculations)

= 1372 J

Problem 15 (5 marks)

For one second 8 litres drop, 1L water = 1 kg of water m = 8 kg (1 mark)

h = 1.5 metres

P = E/t = here time (t) is 1 sec (1/2 mark)

E = gravitational potential energy = mgh

E = 8(9.8)(1.5) (1 mark for calculation)

= 107.6 J (1 mark for answer)

P= 107.6 /1 (1/2 mark)

= 107.6 W (1 mark)

Problem 12 (4 marks)

m = 70kg

v =10 m/s

Here kinetic energy should equal gravitational potential energy

KE = ½ (70) (10)(10)

= 3500 J

PE = 70(9.8)(h) =3500

h = 50/9.8

h = 5.1 m

(1 mark for showing each calculation, 1 mark for each answer with correct units)

  1. Chapter 3 Question 4

Energy conservation is trying to use less energy to do the same job (1 mark )

½ mark for each of two examples of this

Principle of conservation of energy is that energy is neither created nor destroyed, it only changes form (1 mark)

½ mark for each of two examples

Problem 5 (4 marks)

If a coal burning plant burns 2 tons of coal to generate 6000 kWh of electricity calculate efficiency of the plant as the ratio of electricity to fuel energy input.

From table 3.4 1 ton coal produces 25 x 106 Btu

1kWh = 3413 Btu

2 tons of coal then = 2x 25 x 106 Btu = 50 x 106 Btu ( 1 mark)

in kWh = 50 x 106 /3413 = 14650 kWh ( 1 mark – could also have converted to Btu here and done the efficiency calculation in Btu)

Efficiency = energy output /energy input (calculation 1 mark)

= (6000 kWh)/(14650kWh)

= 41% or 0.41(final answer in either form 1 mark)

  1. Chapter 4

Question 14 (2 marks)

Since the carnot efficiency of a heat engine depends on temperature difference (1 mark) a higher temperature source will be a greater difference that a lower heat source (1 mark)

Problem 11 (4 marks)

A coal fired electrical generating plant has efficiency 38%. Temperature of steam leaving the boiler is 550 C, the temperature of the outside is 20 C. What is the maximum efficiency possible for this plant

This is Carnot efficiency that they are trying to calculate.

Temperatures MUST be in Kelvin 550 C = 823 K 20 C = 293 K

efficiency = (1 – (T(cold))/(T(hot))) *100% (2 marks for calculation)

= (1 – (293/823)) *100%

= (1- 0.64) * 100%

= 64% (or 0.64 for a ratio would also be acceptable though is better to express as a percentage) (1 mark)

As a percentage of the total = 38/64

= 59% or 0.59 (1 mark)

Problem 13 (5 marks)

Would you be willing to back an inventor who is marketing a device that takes in 25,000J of heat at 600K and expels heat at 300K and does 12,000 J of work?

This is a question of efficiencies

First the overall efficiency

Efficiency = useful energy/energy input

Energy input = 25 000J

useful energy = 12 000

Efficiency = 12000/25000 (1 mark)

= 0.48 or 48% (1 mark)

Carnot efficiency

This is a question of Carnot efficiency so use, remember to convert the temperature to Kelvin

efficiency = (1 – (T(cold))/(T(hot))) *100% (1 marks for calculation)

= (1 – (300/600)) *100%

= (1- 0.50) * 100%

= 50% (1 mark)

The actual efficiency is very close to the carnot efficiency so this would be a very energy efficient invention (1 mark)

Additional Problems

  1. How much energy (heat) would be needed to raise 8 kg of water from 25 C to 50 C? (Specific heat capacity of water is 4186 J/(kg*C). (2 marks - one for calculation, one for final answer)

Heat = mcT

T = 50 – 25 = 25 C (or K) ( 1 mark- this could also be embedded in the next line

m= 2 kg

heat = 2 x 4186 x ( 25) ( 1 mark)

heat = 209300 J or 209 kJ (1 mark)

  1. For the chemical reaction ethanol + oxygen = water + carbon dioxide, calculate the energy of this reaction from the bond energies. (and additional diagram will be posted to the assignment page for this question) (6 marks)

Bond tallies

Left side 5 (C-H) +1 (C-C)+ 1 (C-O) + 1 (O-H) + 3 (O=O) (1 mark)

Right Side 6 (O-H) + 4 (C=O) (1 mark)

Energies from the notes

Left side 5 (413)+(346) + (358) + (463) + 3 (498)= 4726kJ (1 mark)

Right side 6 (463) + 4(749) = (1 mark)

Energy of reaction = right side – left side

= 5774- 4726 (1 mark for calculation)

= 1048 kJ (1 mark for final answer)

  1. An engine performs 5500J of work is 11 s. What is the power output in kilowatts?(2 marks - one for calculation, one for final answer)

P = E/t

= 5500 J/ 11s

= 500 W = 0.5 kW