Story 1 l En Halba who now has

Story 1 l En Halba who now has a microwave oven puts in two bowls of soup. After heating for one minute, he finds that the bowl with more soup is just right for drinking but the bowl with small amount of soup is too hot. Why is it that the two bowls of soup have different temperatures?

Story 2 l Pn Halba is preparing the ingredients to make some bread. She puts equal masses of water and cooking oil into the oven and turns on the heat for 2 minutes. She discovers that the oil became very hot but the water was just warm to touch. Why the oil hotter than the water?

SPECIFIC HEAT CAPACITY l At the end of the lesson, you should be able to: l Define specific heat capacity l State that l Describe applications of specific heat capacity l Solve problems involving specific heat capacity.

Explanation for story 1 l The same amount of heat was supplied to the soup in each bowl. l The bigger mass of soup is said to have a larger heat capacity because it experiences a smaller temperature increase. l This means that a larger amount of heat would be required to raise the temperature of the bigger mass of soup to the same temperature as the small mass of soup.

Heat capacity l The heat capacity of an object is the amount of heat required to increase the temperature of the object by 1°C. l Heat capacity is expressed in unit J°C-1 or JK-1. l Heat capacity denoted as C (capital C). l Suppose the water in a kettle has a heat capacity of 9000 J°C-1. This means that when the water absorbs 9000 J of heat energy, its temperature will increase by 1°C.

l Heat capacity of an object is calculated in the following way: Heat capacity = heat absorbed or released by object change in temperature l Heat capacity of an object depends on the: l Mass of the object l Types of material l The amount of heat supplied l An object with a larger mass will have a larger heat capacity than an object with smaller mass of the same material.

Explanation for story 2 l The water and the oil were supplied with heat at the same rate. l The oil has a smaller heat capacity than the water since it experiences a larger increase in temperature. l Since both the oil and water have equal masses, the oil is said to have a smaller specific heat capacity than water.

l When comparing the heat capacities of different substances, we refer to their specific heat capacities. The term “specific” means one unit of mass, that is 1 kg of mass when SI units are used.

Specific heat capacity l The specific heat capacity of a substance is the amount of heat that must be supplied to increase the temperature by 1°C for a mass of 1 kg of the substance. l Specific heat capacity is expressed with units Jkg-1°C-1 or Jkg-1 K-1. l The specific heat capacity of aluminum is 900 Jkg-1°C-1. This means that 900 J of heat required to raise the temperature of a 1 kg block of aluminum through 1°C.

l Specific heat capacity, c can be calculated from the amount of heat supplied, Q to a mass, m of a substance and the change in temperature, Ư. Specific heat capacity: l Therefore, the quantity of heat absorbed or lost from a body is given by: Q=mcƯ Heat absorbed Heat lost increase in temperature decrease in temperature

l Other forms of energy such as electrical energy, potential energy and kinetic energy can be converted to heat energy. Heater Electrical energy Power , P Heat energy Pt = mcØ Potential energy Object falls from a high position Heat energy Kinetic energy Moving object stopped due to friction Heat energy mgh = mcØ ½ mv 2 = mcØ

List of specific heat capacities for some common substances

l When two objects of equal mass are heated at equal rates, the object with the smaller specific heat capacity will have a faster temperature increase. l When two hot objects of equal mass are left to cool down, the temperature of the object with smaller heat capacity will drop faster.

Applications of specific heat capacity Cooking pot l The cooling system of car engine l Sea breeze l Land breeze l

Problem solving…. .

1. How much heat energy is required to raise the temperature of a 4 kg iron bar from 32°C to 52°C? [ Specific heat capacity of iron = 452 Jkg-1°C-1] 2. A bottle containing 1. 5 kg of water at 34°C is put into a refrigerator. What is the temperature of the water after 1. 575 x 105 J of heat has been removed from the water? [Specific heat capacity of water = 4200 Jkg-1°C-1] 3. An electric kettle has a power rating of 2. 4 k. W. How long does it take the kettle to heat up 4. 5 kg of tap water from 28°C to 100°C? [Specific heat capacity of water = 4200 Jkg-1°C-1]

Solution 1. Amount of heat required, Q = mcØ = 4 kg x 452 Jkg-1°C-1 x (52 -32)°C = 36160 J Q = mcØ 1. 575 x 105 J = 1. 5 kg x 4200 Jkg-1°C-1 x (34 – t ) °C t = 9°C 2.

Q = mcØ Pt = mcØ 2. 4 k. W x t = 4. 5 kg x 4200 Jkg-1°C-1 x (10028) °C t = 567 s 3.