When you heat a substance, you are transferring energy into it by placing it in contact with surroundings that have a higher temperature

INTRODUCTION

P-26 SC TEAM

OBJECTIVES

define and distinguish between various units of heat

define the mechanical equivalent of heat

discuss everyday examples to illustrate these concepts

Heat is energy that flows from a higher temperature object to a lower one because of difference in temperature.

Heat can change the state of substances.

When heated

WHAT IS HEAT?

UNIT OF HEAT

Heat is measured in unit of calorie. One calorie is defined as the amount of the

heat energy required to raise the temperature of 1ºC of 1 gram of water.

It will take 500 calories to heat 500 grams of water from 20ºC to 21ºC.

In SI, heat is measured in unit of joule. One calorie is equal to 4,184 joule, and it is frequently rounded up to 4,2 joule.

Unit of HEAT

Heat is measured in Joule (J) or calorie (cal).

1 Joule = 0,24 calorie 1 calorie = 4,186 Joule.

If an object has become hotter, it means that it has gained heat energy.

If an object cools down, it means it has lost energy

1 kilo calorie = 1 kcal = 4.186 Joule.

MECHANICAL EQUIVALENT OF HEAT

Joule demonstrated that water can be heated by doing (mechanical) work, and showed that for every 4186 J of work done, the temperature of water raise by 1 C⁰ per kg.

Q = m c T The temperatureChange (C)

Specific heat capacity (J / kg C)

Mass of object (kg)

Thermal energy (J)

ΔT = T final - Tinitial

THE THERMAL ENERGY (Q)

Greater amount of heat is required to raise the temperature of greater mass of the material.The amount of thermal energy is influenced by:

Mass

Type of object

Temperature change

The bigger mass the bigger thermal energy.

The difference object has difference thermal energy.

The higher temperature change, the higher thermal energy.

THE AMOUNT OF THERMAL ENERGY (Q)

If ΔT is positive, Q is also positive. It means that a substance has an increase in temperature and gets the heat energy (receives the heat).

If ΔT is negative, Q is also negative; it means that the substance has a decrease in temperature and will lose the heat energy (releases the heat).

What is the amount of heat required to raise the temperature of 2 kg of copper from 30 C to 80 C, given the specific heat capacity of cooper is 400 J / kg C ?

Solution:m = 2 kgT = 80 C – 30 C = 50 C c = 400 J / kg C

Q = m c T = 2 kg x 400 J / kg C x 50 C = 40 000 J

So, the amount of heat is 40 000 J

EXAMPLE

A silver spoon has a mass of 32 g and then it is cooled from 60ºC down to 20ºC. How much heat is released by the spoon?given the specific heat capacity of silver is 235 J/(kgºC)

EXAMPLE

A silver spoon has a mass of 32 g and then it is cooled from 60ºC down to 20ºC. How much heat is released by the spoon?

Known:m = 32 g = 0.032 kgc = 235 J/(kgºC) TInitial = 60ºC

Tfinal = 20ºC

ΔT = Tf – Ti = -40ºC

Question Q = ..... ?

Solution: Q = m . c . Δ T

= 0.032 kg. 235 J/(kgºC). -40ºC

= - 301 Jspoon releases the heat of 301 J. So that the spoon becomes cooler.

Heat Capacity (C)

Heat capacity (C) is defined as the amount of thermal energy required to raise the temperature of body by 1 K or 1 C.

Temperature change by T C

C = QT

Specific Heat Capacity (c)

The amount of thermal energy needed by 1 kilogram of substance to increase its temperature up to 1 Kelvin is called specific heat capacity

Temperature changes byT C

c = Qm T

m kg

SPECIFIC HEAT CAPACITY OF SOME MATERIALS

Material Specific Heat capacity

(J / kg C )

Material Specific Heat capacity

(J / kg C )

AluminumBrassCooperIceIronLead

920380400

2100460130

SteamMercuryMethylated spirit Sea waterWaterZinc

2010140240039004200390

Water and other substances that have high specific heat capacity can absorb more heat energy with less change of temperature.

Latent heat of fusion, (Lf) is heat per unit mass needed to change a substance from solid to liquid without change in temperature.

Latent heat of vaporization, (LV) is heat per unit mass needed to change a substance from liquid to solid without change in temperature.

Lf =Qm

LV =Qm

Q = m Lf

Q = m LV

LATENT HEAT (L) AND SPECIFIC LATENT HEAT

PHASES OF MATTER Heat required for phase changes:

Vaporization: liquid vapour Melting: solid liquid Sublimation: solid vapour

Heat released by phase changes: Condensation: vapour liquid Fusion: liquid solid Deposition: vapour solid

PHASES OF MATTER

GRAPH OF PHASE CHANGES

Useful data for thermal energy calculation

Specific latent heat of water = 336 000 J/kg

Specific latent heat of steam = 2 260 000 J/kg

Specific heat capacity of water = 4200 J/kg C

Specific heat capacity of ice = 2100 J/kg C

Specific heat capacity of water = 1 cal/gram C

Specific heat capacity of ice = 0,5 cal/gram C

Example:

What is the amount of thermal energy required to change the state of 1 gram of ice at – 30 C to steam at 120 C?

Energy added (J)

T (C)

- 30

0

100

120

Water + steam

WaterIce + Water

Ice

Steam

A

B

C

DE

Solution:The figure below indicates the experimental results obtained when energy is gradually added to the ice. Let us examine each portion of the red curve.

Part A:On this portion of the curve, the temperature of the ice changes from- 30. °C to 0°C. The specific heat of ice is 2100 J / kg C.

Q = m ice c ice T = (1 x 10 -3 kg) (2100 J/kg C) (0 C – (-30 C)) = 63 J

Part B:When the temperature of the ice reaches 0.0°C, the ice–water mixture remains at this temperature - even though energy is being added - until all the ice melts.

Q = m ice Lf

= (1 x 10-3 kg) (336 000 J/kg) = 336 J

Part C:Between 0.0°C and 100.0°C, nothing surprising happens. No phase change occurs, and so all energy added to the water is used to increase its temperature. The amount of energy necessary to increase the temperature from 0.0°C to 100.0°C is:

Q = m water c water T

= (1 x 10-3 kg) ( 4 200 J/kg °C) (100 °C)

= 420 J

Part D:At 100.0°C, another phase change occurs as the water changes from water at 100.0°C to steam at 100.0°C.

Q = m water LV

= (1 x 10-3 kg) (2 260 000 J/kg)

= 2 260 J

Part E:On this portion of the curve, as in parts A and C, no phase change occurs; thus, all energy added is used to increase the temperature of the steam. The energy that must be added to raise the temperature of the steam from 100.0°C to 120.0°C is

Q = m steam c steam T = (1 x 10-3 kg) (2 010 J/kg °C) (120 – 100 °C)

= (1 x 10-3 kg) (2 010 J/kg °C) (20 °C) = 40.2 J

The total amount of energy that must be added to change 1 gram of ice at - 30.0°C to steam at 120 0°C is the sum of the results from all five parts of the curve.

Q = QA + QB + QC + QD + QE

= 63 J + 336 J + 420 J + 2 260 J + 40.2 J = 3119.2 J

CALORIMETERS

BLACK’S PRINCIPLE

BLACK’S PRINCIPLE