In the previous chapter, we completed a discussion on mechanical properties of fluids. In this chapter we will see thermal properties of matter.
First we will see some basics about temperature. It can be written in 4 steps:
1. Normally, when we touch a body, we get a sense about whether it is hot or cold
• We may even be able to say whether the body is too hot or too cold
2. But such an assessment of heat will not be accurate enough for scientific and engineering purposes
• For example, a body may need to have an exact amount of ‘hotness’ in order to use it in an experiment
♦ A higher hotness or lower hotness may not be allowable
• In the same way, a body may need to have an exact amount of ‘coldness’ in order to use it in an experiment
♦ A higher coldness or lower coldness may not be allowable
3. So scientists developed the concept of 'temperature'
• Temperature enables us to measure the exact hotness or coldness of a body
♦ A hot body will give a higher temperature reading
♦ A cold body will give a lower temperature reading
• Even minute differences in hotness or coldness can be recorded by taking accurate temperature readings
4. Temperature is a physical quantity just like length, area, mass etc.,
This can be explained in 3 steps:
(i) Consider the length of two ropes
• We can say one rope is longer than the other
• We can be more specific if we measure the length of each rope
• For that, we take measurements in cm or inch
♦ cm and inch are two different scales for measuring lengths
(ii) Consider the area of two rooms
• We can say one room is larger than the other
• We can be more specific if we measure the area of each room
• For that, we take measurements in m2 or sq.feet
♦ m2 and sq.ft are two different scales for measuring areas
(iii) In the same way, we can write about temperature
• We can say one object is hotter than another object
• We can be more specific if we measure the temperature of each object
• For that, we take measurements in oC, oF or K
• oC (degree Celsius), oF (degree Fahrenheit) and K (degree Kelvin) are three different scales for measuring temperatures
• We will see the details about the three scales later in this section
1. Consider a glass of ice-cold water kept open to the atmosphere, on a hot summer day
• We would find that, as time passes, the water becomes less and less cold
• This is because, heat from the surroundings flows into the glass of water
• Heat is a form of energy. A gain in this energy, causes the water to warm up
2. Consider a cup of hot tea kept open to the atmosphere, on the same hot summer day
• We would find that, as time passes, the tea becomes less and less hot
• This is because, heat from the cup flows out into the surroundings
• Heat is a form of energy. A loss in this energy, causes the tea to cool down
3. We must always remember an important point. It can be written in two steps:
(i) In the case of cold water, the flow of heat took place because, there was a temperature difference between the water and the surroundings
• If there was no temperature difference, the water would have remained at the same temperature
(ii) In the case of the hot tea, the flow of heat took place because, there was a temperature difference between the tea and the surroundings
• If there was no temperature difference, the tea would have remained at the same temperature
4. So we can write the definition of heat. It can be written in 3 steps:
(i) Heat is a form of energy
(ii) Transfer of this energy can take place in two different situations:
♦ Transfer between two or more systems
♦ Transfer between a system and it's surroundings
(iii) A transfer will take place only if there is a temperature difference
Next we will see various methods to measure temperature. The basics can be written in 4 steps:
1. The device used to measure temperature is called thermometer
2. Usually liquids like mercury or alcohol is used in glass thermometer
• When heat is applied, mercury expands. That is., volume of mercury increases
• So it rises out from it’s container into a vertical tube
• The container is in the form of a bulb
3. Let the temperature at the bulb be such that, water just freezes
• ‘Just freeze’ indicate that, the temperature is just enough to cause the water to freeze
♦ A higher temperature will cause the frozen water to melt
♦ A lower temperature is not required necessary
• In such a situation, we put a mark at the level of mercury in the vertical tube
♦ At that mark, we write: 0 oC
♦ If it is a Fahrenheit thermometer, we write: 32 oF
♦ If it is a kelvin thermometer, we write: 273 K
4. Let the temperature at the bulb be such that, water just starts to boil
• ‘Just boil’ indicate that, the temperature is just enough to cause the water to boil
♦ A higher temperature is not necessary for boiling
♦ A lower temperature will not make the water boil
• In such a situation, we put a mark at the level of mercury in the vertical tube
♦ At that mark, we write: 100 oC
♦ If it is a Fahrenheit thermometer, we write: 212 oF
♦ If it is a kelvin thermometer, we write: 373 K
• The basics about the three scales and their interrelations can be seen here and here
• Some solved examples related to various temperature scales are given at the end of this chapter in section 11.10
1. Most substances expand when heat is applied. Also they contract when cooled
■ A change in temperature of a body causes change in it’s dimensions. This is called thermal expansion
♦ The expansion in length is called linear expansion
♦ The expansion in area is called area expansion
♦ The expansion in volume is called volume expansion
• Let us see each one in detail
2. First we will see linear expansion
• This can be explained in 5 steps:
(i) In fig.11.1(a) below,
♦ When the temperature is t1, length of the rod is l1
♦ When the temperature is t2, length of the rod is l2
✰ Then change in length (Δl) = l2-l1
✰ Also change in temperature (Δt) = t2-t1
 |
| Fig.11.1 |
♦ The original length l1
♦ The change in temperature Δt
(iii) That means:
♦ If the original length is greater, we will get a greater Δl
♦ If the change in temperature is greater, then also, we will get a greater Δl
(iv) So we can write: Δl ∝ (l1 × Δt)
• Thus we get Eq.11.1: Δl = (𝛼L × l1 × Δt)
♦ Where 𝛼L is the constant of proportionality
✰ It is called: coefficient of linear expansion
(v) We have: $\mathbf\small{\rm{\alpha_L=\frac{\Delta L}{l_1 \times \Delta t}}}$
• So the unit of 𝛼L can be obtained as: $\mathbf\small{\rm{\frac{(m)}{(m) \times (K)}=K^{-1}}}$
3. Next we will see area expansion
• This can be explained in 5 steps:
(i) In fig.11.1(b) above,
♦ When the temperature is t1, area of the plate is A1
♦ When the temperature is t2, area of the plate is A2
✰ Then change in area (ΔA) = A2-A1
✰ Also change in temperature (Δt) = t2-t1
(ii) The change in area (ΔA) is found to be directly proportional to two items:
♦ The original area A1
♦ The change in temperature Δt
(iii) That means:
♦ If the original area is greater, we will get a greater ΔA
♦ If the change in temperature is greater, then also, we will get a greater ΔA
(iv) So we can write: ΔA ∝ (A1 × Δt)
• Thus we get Eq.11.3: ΔA = (𝛼A × A1 × Δt)
♦ Where 𝛼A is the constant of proportionality
✰ It is called: coefficient of area expansion
(v) We have: $\mathbf\small{\rm{\alpha_A=\frac{\Delta A}{A_1 \times \Delta t}}}$
• So the unit of 𝛼A can be obtained as: $\mathbf\small{\rm{\frac{(m^2)}{(m^2) \times (K)}=K^{-1}}}$
4. Next we will see volume expansion
• This can be explained in 5 steps:
(i) In fig.11.1(c) above,
♦ When the temperature is t1, volume of the block is V1
♦ When the temperature is t2, volume of the block is V2
✰ Then change in volume (ΔV) = V2-V1
✰ Also change in temperature (Δt) = t2-t1
(ii) The change in volume (ΔV) is found to be directly proportional to two items:
♦ The original volume V1
♦ The change in temperature Δt
(iii) That means:
♦ If the original volume is greater, we will get a greater ΔV
♦ If the change in temperature is greater, then also, we will get a greater ΔV
(iv) So we can write: ΔV ∝ (V1 × Δt)
• Thus we get Eq.11.2: ΔV = (𝛼V × V1 × Δt)
♦ Where 𝛼V is the constant of proportionality
✰ It is called: coefficient of volume expansion
(v) We have: $\mathbf\small{\rm{\alpha_V=\frac{\Delta V}{V_1 \times \Delta t}}}$
• So the unit of 𝛼V can be obtained as: $\mathbf\small{\rm{\frac{(m^3)}{(m^3) \times (K)}=K^{-1}}}$
1. Consider two objects:
♦ a glass rod (made with pyrex glass)
♦ a copper rod
• Both of them have the same length L
2. Let both of them be heated through the same change in temperature Δt
3. Then we can write:
♦ Change in length suffered by glass (ΔL(glass)) = (𝛼L(glass) × L × Δt)
♦ Change in length suffered by copper (ΔL(copper)) = (𝛼L(copper) × L × Δt)
4. From the data book, we have:
♦ 𝛼L(glass) = 0.32 × 10-5 K-1
♦ 𝛼L(copper) = 1.7 × 10-5 K-1
5. Substituting the values in (3), we get:
♦ ΔL(glass) = (0.32 × 10-5 × L × Δt)
♦ ΔL(copper) = (1.7 × 10-5 × L × Δt)
6. Taking ratios, we get:
$\mathbf\small{\rm{\frac{\Delta L_{(copper)}}{\Delta L_{(glass)}}=\frac{1.7}{0.32}}}$ = 5.31
• That means:
♦ For the same rise in temperature,
♦ linear expansion of copper will be about five times that of glass
1. Consider two samples:
♦ a sample of mercury
♦ a sample of alcohol
• Both samples have the same volume V
2. Let both samples be heated through the same change in temperature Δt
3. Then we can write:
♦ Change in volume suffered by mercury (ΔV(mercury)) = (𝛼V(mercury) × V × Δt)
♦ Change in volume suffered by alcohol (ΔV(alcohol)) = (𝛼V(alcohol) × V × Δt)
4. From the data book, we have:
♦ 𝛼V(mercury) = 18.2 × 10-5 K-1
♦ 𝛼V(alcohol) = 110 × 10-5 K-1
5. Substituting the values in (3), we get:
♦ ΔV(mercury) = (18.2 × 10-5 × V × Δt)
♦ ΔV(alcohol) = (110 × 10-5 × V × Δt)
6. Taking ratio, we get:
$\mathbf\small{\rm{\frac{\Delta V_{(mercury)}}{\Delta V_{(alcohol)}}=\frac{110}{18.2}}}$ = 6.04
• That means:
♦ For the same rise in temperature
♦ volume expansion of alcohol will be about six times that of mercury
In the next section, we will see more details about volume expansion
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