Types of energy POTENTIAL ENERGY : STORED ENERGY. The energy inside the substance. KINETIC ENERGY : Associated with motion. Average KE = TEMPERATURE.

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Presentation transcript:

Types of energy POTENTIAL ENERGY : STORED ENERGY. The energy inside the substance. KINETIC ENERGY : Associated with motion. Average KE = TEMPERATURE

HEAT OF FUSION FOR WATER (TABLE B) Amount of heat needed to completely melt 1g of water (ice!). 334 J/g 334 Joules of heat are necessary to completely melt 1 g of water. HOW MUCH HEAT IS NEEDED TO MELT 10 g OF WATER?

HEAT OF VAPORIZATION FOR WATER (TABLE B) The amount of heat needed to completely vaporize one g of water at its boiling point J/g Water needs 2260 J of heat per gram to convert to gas!

When there is a change in temperature use Q = m x C x  T While the substance is melting/freezing use Q = m x H f While the substance is boiling/condensing use Q = m x H vap

How much heat is needed to completely melt 10 g of ice at 0 0 C ? How much heat is needed to vaporize 10 g of water at C ?

Energy Changes Associated with Changes of State

Melting Point and Freezing Point The temperature at which a substance melts. Is the same temperature at which the substance freeze. Boiling Point and Condensation Point are the same temperature. Normal boiling point is the temperature at which a liquid boils at normal pressure.

1.-Calculate the amount of heat needed to increase the temperature of 200g of water from 0 0 C to C. Specific heat capacity of water = 4.18J

2.-How much heat is needed to completely vaporize 200 g of water at C

A 200g sample of metal with a specific heat of 0.50 J/gC is heated to 90.0C and then placed in a g sample of water at 30.0C. What is the final temperature of the metal and the water?

Heat gained by one object must equal heat lost by another

Heat transfer Therefore the final temperature of both objects will be the same but one object will lose heat and the other will gain heat But the heat energy gained must equal the heat energy lost Therefore q lost must equal q gained

A 200g sample of metal with a specific heat of 0.50 J/gC is heated to 90.0C and then placed in a g sample of water at 30.0C. What is the final temperature of the metal and the water?

Heat energy lost by one system = heat energy gained by another system q metal = q H 2 O MC∆T = MC∆T 200g. 0.5 (T - x ) = (x – T) 200g. 0.5 (90 - x ) = (x – 30) 100(90 –x) = 418 (x-30) 9000 – 100x = 418x – = 418x + 100x = 518x divide o C = X where x = Tf

A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water? Heat energy lost by one system = heat energy gained by another system q metal = q H 2 O MC∆T = MC∆T

A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water? Metal Water Heat energy lost by one system = heat energy gained by another system q metal = q H 2 O MC∆T = MC∆T 1000 g. 0.5 (T - x ) = (x – T)

Mass of metal 1 kg = 100g Specific heat capacity = 05 Since the metal is losing energy and its temperature will decrease Specific heat capacity f or water = 4.18 Since the final temp of the water will be higher than the initial temp

A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water? Metal Water Heat energy lost by one system = heat energy gained by another system q metal = q H 2 O MC∆T = MC∆T 1000 g. 0.5 (T - x ) = (x – T) 500 (100-x) = 209 (x-20) – 500x = 209x – = 709x = x = x final temp o C