+ All Categories
Home > Documents > Electricity & heat energy

Electricity & heat energy

Date post: 26-Jun-2015
Category:
Upload: nmacintoshwqsbqcca
View: 483 times
Download: 0 times
Share this document with a friend
Popular Tags:
20
Module 2 – Heat Energy Objectives 5.2 Calculate heat energy, using a mathematical relationship.
Transcript
Page 1: Electricity & heat energy

Module 2 – Heat Energy

Objectives5.2 Calculate heat energy, using a

mathematical relationship.

Page 2: Electricity & heat energy

Heat Energy:

Heat (thermal) energy is simply a type of energy Ex. heat from a stove Ex. heat from a light bulb

Heat energy can be transferred from one object to another Ex. from a hot coffee cup to your

cold hands

What is heat energy?

Page 3: Electricity & heat energy

Q = mC∆T

Heat Energy

Joules (J)

mass

grams (g)

Change in

Temperature

Degrees

oC

Specific Heat

Capacity

(J/g•oC)

Page 4: Electricity & heat energy

Specific Capacity (c)

specific heat capacity (c) - the amount of energy absorbed by a substance or object.

It is a characteristic property.

Substances Joules/(g• 0C)

Copper 0.383

Iron 0.452

Aluminum 0.896

Antifreeze 2.2

Methanol 2.547

Water 4.19

Every substance has their own specific heat capacity that does not change. However, water is most commonly used in exam

questions.

Page 5: Electricity & heat energy

A pot with 500g of water is placed a stove. The temperature of the water increased from 20oC to 100oC. How much heat was absorbed by the water?

m = 500g

Ti = 20oC

Tf = 100oC

c = 4.19 J/goC

Q = mc∆T

Q = (500)(4.19)(100-20)

Q = 167,600 J

Recall: The “c” of water is 4.19 J/goC

Page 6: Electricity & heat energy

A 250g glass of cold ice water is placed outside in the sun. The temperature of the water rises from 5oC to 25oC. How much heat was absorbed by the water?

m = 250g

Ti = 5oC

Tf = 25oC

c = 4.19 J/goC

Q = mc∆T

Q = (250)(4.19)(25-5)

Q = 20,950 J

Page 7: Electricity & heat energy

A 20g iron frying pan is heated from 21oC to 85oC. If the specific heat capacity of iron is 0.45 J/goC, how much heat was absorbed by the firing pan?

m = 20g

Ti = 21oC

Tf = 85oC

c = 0.45 J/goC

Q = mc∆T

Q = (20)(0.45)(85-21)

Q = 576 J

Page 8: Electricity & heat energy

A 250mL of water at 30oC is heated until it boils at 100oC. How much heat was absorbed by the water?

m = 250 g

Ti = 30oC

Tf = 100oC

c = 4.19 J/goC

Q = mc∆T

Q = (250)(4.19)(100-30)

Q = 73,325 J

To go from volume to mass: times volume of water by the

density of water. So,

1g/mL x 250 mL = 250 g

Note for water: 1mL = 1g

Page 9: Electricity & heat energy

A 1500mL of water at 5oC is heated until it reaches 37oC. How much heat was absorbed by the water?

m = 1500 g

Ti = 5oC

Tf = 37oC

c = 4.19 J/goC

Q = mc∆T

Q = (1500)(4.19)(37-5)

Q = 201,120J

Note for water: 1mL = 1g

Page 10: Electricity & heat energy

20,112J of energy is used to heat water from 2oC to 50oC. How many grams of water was heated?

m = ? g

Ti = 2oC

Tf = 50oC

c = 4.19 J/goC

Q = 20, 112J

Q = mc∆T

20,112 = (?)(4.19)(50-2)

m = 100 g

20,112 = (?)(201.12)

20,112 = (?)(201.12) 201.12 201.12

Page 11: Electricity & heat energy

360J of energy is used to heat a 20g piece of iron from 20oC to 60oC. What is the heat capacity of iron?

m = 20 g

Ti = 20oC

Tf = 60oC

Q = 360J

c = ? J/goC

Q = mc∆T

360 = (20)(?)(60-20)

c = 0.45 J/goC

360 = (800)(?)

360 = (800)(?) 800 800

Page 12: Electricity & heat energy

A calorimeter was used to heat 300ml of water to 70oC. If the thermal (heat) energy was equal to 62850 J, what was the final temperature of the water before being heated?

m = 300 g (remember 1 g = 1ml)

c = 4.19 Joules/(g• 0C)

Ti= ?

TF= 70 0C

Q = 62850 J

Q = mc∆T62850 = 300 x 4.19(70 - Ti)62850 = 1257(70 - Ti)62850 = 87990 – 1257Ti

-25140 = - 1257Ti

20 0C= Ti

TrickyQuestion

Page 13: Electricity & heat energy

A calorimeter was used to heat 400ml of water to 40oC. If the thermal (heat) energy was equal to 5.4 kJ, what was the initial temperature of the water before being heated?

m = 400 g (remember 1 g = 1ml)

c = 4.19 Joules/(g• 0C)

Ti= ?

TF= 40 0C

Q = 5400 (5.4kJ = 5400 J)

Q = mc∆T5400 = 400 x 4.19(40 - Ti)5400 = 1676(40 - Ti)5400 = 67040 – 1676Ti

-61640 = - 1676Ti

36.8= Ti

TrickyQuestion

Page 14: Electricity & heat energy

Q = mc∆T (J) (g)(J/g•oC)

o Heat capacity of water is 4.19 J/g•oC

o 1 g of water = 1 ml of water

o Q is always in Joule (j) not kJ!

Tf –Ti

(oC)

Page 15: Electricity & heat energy

Activities

Worksheet # 17

Page 16: Electricity & heat energy

First Law of Thermodynamics

This law states that “in an isolated system energy remains constant”.

This means that energy can be transformed (neither created nor destroyed).

Example: When you eat a candy bar the bar is transformed into energy within your body.

Page 17: Electricity & heat energy

Different Types of Energy

Thermal

Light

Electrical

Page 18: Electricity & heat energy

Different Types of Energy

Magnetic

Nuclear

Kinetic

Page 19: Electricity & heat energy

Potential

Chemical

Sound

Aeolian

Different Types of Energy

Page 20: Electricity & heat energy

Tidal

Geothermal

Hydro

Solar

Different Types of Energy


Recommended