Heat and the first law of thermodynamics

Heat and the first law of thermodynamics

A thunderstorm is driven by the violent collision of warm air with cooler air.

18.6

18.7 18.1

LATENT HEAT AND PHASE TRANSITIONS

DEFINITION OF HEAT

SPCIFIC HEAT OF SOLIDS AND

FLUIDS Chapter

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Outline

After studying this chapter, you will be able to:

1. Identify heat.

2. Identify the specific heat of various materials of solid and fluid.

3. Explain phase transition.

4. Define the latent heat.

5. Differentiate between the latent heat of fusion and the latent heat of vaporization.

Learning Objectives

First Law of Thermodynamics

This chapter examines the nature of heat.

Heat is a form of energy that is transferred into or out of a system.

Heat is governed by a more general form of the law of conservation of energy, known as the First Law of Thermodynamics.

Heat is essential to life processes; no life could exist on earth without heat from the sun or from the earth’s interior.

1

2

3

4

What is heat ?

 Heat is one of the most common forms of energy in the universe, and we all experience it every day.

 Heat is energy in transfer: to or from an object.

 It’s symbol is Q and it is measured by Joules = J.

 Heat is the reason substances change from:

Solids liquids gases

 Heat always flows from high temperatures to low temperatures.

Definition of Heat

• A cup of water at 20 °C (temperature of sample T_s) is placed in a room on a hot day 30°C (temperature of environment T_e ).

• The cup gains heat energy (Q) and heats up.

• Its temperature increases until it reaches the temperature of the air in the room.

If thermal energy is transferred into a system, then Q > 0.

• A cup of water at 20 °C (temperature of sample T_s) is placed in a room on a cold day 10°C (temperature of environment T_e ).

• The cup loses heat energy (Q) and cools down.

• Its temperature decrease until it reaches the temperature of the air in the room.

If thermal energy is transferred from a system, then Q < 0

Definition of Heat

• A cup of water at 20 °C (temperature of sample T_s) is placed in a room at 20°C (temperature of environment T_e ).

• The cup neither gains not loses heat energy (Q).

• Its temperature remains 20 °C.

If the system and its environment have the same temperature, then Q = 0.

• If T_s ≠ T_e , then the temperature of the system changes until it is equal to the temperature of the environment (thermal equilibrium).

• At thermal equilibrium, the water, the cup, and the air in the room are all at the same temperature.

Definition of Heat

What does a positive value for Q mean?

When does heat transfer between a system and its environment?

What is the meaning of thermal equilibrium?

Definition of Heat

1

2

3

Heat capacity (C):-

Is the amount of heat energy required to increase the temperature of a substance 1°C.

The SI unit of heat capacity is: (J/K) or (J/ °C )

Notice: the abbreviation of heat capacity is capital letter C

Specific Heat of Solids and Fluids

Heat (J)

change in temperature (K) Or (°C)

Heat capacity

C = Q

∆T

Specific heat (c):-

Is the amount of heat energy required to increase the temperature of 1Kg of substance by 1°C.

Where

Q: heat (J)

M: mass of substance (Kg)

∆T: change in temperature (K or °C) which is equal to 𝑇_𝑓 − 𝑇_𝑖, final temperature minus initial temperature

c: specific heat [(J/kg.K) or (J/kg.°C)]

Notice: the abbreviation of heat capacity is small letter c.

Specific Heat of Solids and Fluids

Q = c m ∆T

• The specific heats of various materials are given in the table.

Specific Heat of Solids and Fluids

Q1: You have 2.00 kg of water at a temperature of 20.0 °C. How much energy is required to raise the temperature of that water to 95.0 °C?

Extra Exercise

Ans: The energy required to warm 2.00 kg of water from 20.0 °C to 95.0 °C is:

c_water From table

= 628.5𝑘𝐽

There is no need to convert °C to K because the difference in temperature of both scales are the same. 𝛥T_C = 𝛥T_K

Q2: A certain amount of heat will warm 1g of material A by 3 ˚C and 1g of material B by 4˚C.

Find the ratio between their specific heats?

Extra Exercise

a)

_𝒄^𝒄𝑨

𝑩

=

^𝟑_𝟒

b)

_𝒄^𝒄𝑨

𝑩

=

^𝟒_𝟑

c)

_𝒄^𝒄𝑨

𝑩

= 1

Concept Check 18.1

A. block 1 B. block 2 C. block 3

D. All of the blocks received the same amount of heat.

Suppose you raise the temperature of:

copper block 1 from –10 °C to +10 °C, copper block 2 from +20

°C to +40 °C, and copper block 3 from +90 °C to +110 °C.

Assuming that the blocks have the same mass, to which one did you add the most heat?

Ans: D (as the blocks are all the same material with the same rise in temperature).

Q3: 5.00 kg of some liquid at 10.0C is mixed with 1.00 kg of the same liquid at 40.0C. What is the final equilibrium temperature? Ignore any heat flow between the containers and/or surroundings.

A. 12.0C B. 15.0C C. 18.0C D. 25.0C

Extra Exercise

 SOLUTION:

If the heat gained by 5kg of liquid is + Q₁ and the heat lost by 1kg of liquid is – Q₂.

At thermal equilibrium the total amount of heat is zero so

𝑸_𝟏= − 𝑸_𝟐 or

𝑸_𝟏+𝑸_𝟐 = 𝟎

𝒎_𝟏𝒄_𝟏∆𝑻_𝟏 + 𝒎_𝟐𝒄_𝟐∆𝑻_𝟐 = 𝟎 Same liquid 𝑐₁= 𝑐₂

Sample Problem 18.1

PROBLEM:

A metalsmith pours 3.00 kg of lead shot at a temperature of 94.7 °C into 1.00 kg of water at 27.5 °C in an insulated container.

What is the final temperature of the mixture?

Water and Lead p 242

SOLUTION:

■ The sum of the heat lost by the lead shot and the heat gained by the water is zero, because the process took place in an insulated container and because the total energy is conserved.

Sample Problem 18.1 Water and Lead p 242

𝑚_{𝑙𝑒𝑎𝑑} = 3𝑘𝑔, 𝑇_{𝑖,𝑙𝑒𝑎𝑑} = 94.7℃, 𝑇_{𝑓,𝑙𝑒𝑎𝑑} = 𝑇, 𝑐_{𝑙𝑒𝑎𝑑} = 0.129𝑘𝐽/(𝑘𝑔𝐾) 𝑄 = 𝑐 𝑚∆𝑻

𝑄_{𝑙𝑒𝑎𝑑} = 𝑐_{𝑙𝑒𝑎𝑑} 𝑚_{𝑙𝑒𝑎𝑑}∆𝑻_{𝒍𝒆𝒂𝒅}

= 0.129 × 3 × (𝑇 − 94.7) = 0.387(𝑇 − 94.7)

𝑚_{𝑤𝑎𝑡𝑒𝑟} = 1𝑘𝑔, 𝑇_{𝑖,𝑤𝑎𝑡𝑒𝑟} = 27.5℃, 𝑇_{𝑓,𝑤𝑎𝑡𝑒𝑟} = 𝑇, 𝑐_{𝑤𝑎𝑡𝑒𝑟} = 4.19 𝑘𝐽/(𝑘𝑔𝐾) 𝑄_{𝑤𝑎𝑡𝑒𝑟} = 𝑐_{𝑤𝑎𝑡𝑒𝑟} 𝑚_{𝑤𝑎𝑡𝑒𝑟}∆𝑻_{𝒘𝒂𝒕𝒆𝒓}

= 4.19 × 1 × (𝑇 − 27.5) = 4.19(𝑇 − 27.5)

At thermal equilibrium the total amount of heat is zero

The three common states of matter are solid, liquid, and gas.

• If enough heat is added to a solid, it melts into a liquid.

• If enough heat is added to a liquid, it vaporizes into a gas.

• These are phase changes, or phase transitions.

Latent Heat and Phase Transition

-3C -1C

After some time After some time

0C 𝑇_{𝑚𝑒𝑙𝑡𝑖𝑛𝑔}

0C

After some time

Why is the temperature is not changing while we continue giving heat?

Where did the heat that was given to the ice go?

Where is the heat getting hidden?

What will happen if we continue heating?

0C

After some time 22C

After some time

60C

100C

Why is the temperature not changing while we continue giving heat??

Where does the heat that was given to the water go?

Where is the heat getting hidden?

What will happen if we continue heating?

After some time After some time

100C 𝑇_{𝑏𝑜𝑖𝑙𝑖𝑛𝑔}

100C

Latent heat (L):-

It is the quantity of heat released or absorbed when a substance changes its physical phase at constant temperature.

0 °C 100 °C

100 °C

Latent Heat and Phase Transition

The SI unit of latent heat is (J/Kg).

Latent heat of vaporization, L

vaporization

Latent Heat

Latent heat of fusion, L

_fusion

A change of phase of a

substance from solid to liquid at the same temperature.

A change of phase of a substance from liquid to gas at same

temperature.

The latent heat of fusion for a given substance is different from the latent heat of vaporization for that substance.

Latent Heat and Phase Transition

Latent Heat and Phase Transition

• Latent heat of various materials are given in the table.

How much heat, Q, is required to change the phase of 0.5 kg of water at 100 °C from liquid to gas?

Hint: The Latent heat of vaporization of water is 2260 kJ/kg.

Sample Problem

The heat required for vaporization at 100 °C is:

The END OF

CHAPTER

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