Physics CALORIMETRY, CHANGE OF STATE AND LATENT HEAT

CALORIMETRY
CHANGE OF STATE
LATENT HEAT

### SPECIFIC HEAT CAPACITY

Take some water in a vessel and start heating it on a burner. Soon you will notice that bubbles begin to move upward. As the temperature is raised the motion of water particles increases till it becomes turbulent as water starts boiling.

What are the factors on which the quantity of heat required to raise the temperature of a substance depend? In order to answer this
question in the first step, heat a given quantity of water to raise its temperature by, say 20 °C and note the time taken.

Again take the same amount of water and raise its temperature by 40 °C using the same source of heat.

"Note" the time taken by using a stopwatch. You will find it takes about twice the time and therefore, double the quantity of heat required raising twice the temperature of same amount of water.

In the second step, now suppose you take double the amount of water and heat it, using the same heating arrangement, to raise the
temperature by 20 °C, you will find the time taken is again twice that required in the first step.

In the third step, in place of water, now heat the same quantity of some oil, say mustard oil, and raise the temperature again by 20 °C. Now note the time by the same stopwatch. You will find the time taken will be shorter and therefore, the quantity of heat required would be less than that required by the same amount of water for the same rise in temperature.

The above observations show that the quantity of heat required to warm a given substance depends on its mass, m, the change in
temperature, ΔT and the nature of substance.

The change in temperature of a substance, when a given quantity of heat is absorbed or rejected by it, is characterised by a quantity called the heat capacity of that substance. We define heat capacity, S of a substance

S = (DeltaQ)/(DeltaT).....(11.10)

where ΔQ is the amount of heat supplied to the substance to change its temperature from T to T + ΔT.

You have observed that if equal amount of heat is added to equal masses of different substances, the resulting temperature changes
will not be the same.

It implies that every substance has a unique value for the amount of heat absorbed or rejected to change the
temperature of unit mass of it by one unit. This quantity is referred to as the "specific heat capacity" of the substance

If ΔQ stands for the amount of heat absorbed or rejected by a substance of mass m when it undergoes a temperature change ΔT, then the specific heat capacity, of that substance is given by

s =S/M = 1/m (ΔQ)/(ΔT)......(11.11)

The specific heat capacity is the property of the substance which determines the change in the temperature of the substance (undergoing no phase change) when a given quantity of heat is absorbed (or rejected) by it. It is defined as the amount of heat per unit mass absorbed or rejected by the substance to change its temperature by one unit.

It depends on the nature of the substance and its temperature. The SI unit of specific heat capacity is J kg^(–1) K^(–1).
If the amount of substance is specified in terms of moles μ, instead of mass m in kg, we can define heat capacity per mole of the
substance by

C = S/mu = 1/mu (ΔQ)/(ΔT)......(11.12)

where C is known as molar specific heat capacity of the substance.

Like S, C also depends on the nature of the substance and its temperature.

The SI unit of molar specific heat capacity is J mol^(–1) K^(–1).

However, in connection with specific heat capacity of gases, additional conditions may be needed to define C. In this case, heat transfer
can be achieved by keeping either pressure or volume constant.

If the gas is held under constant pressure during the heat transfer, then it is called the molar specific heat capacity at constant pressure and is denoted by C_p.

On the other hand, if the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume and is denoted by C_v.

Table 11.3 lists measured specific heat capacity of some substances at atmospheric pressure and ordinary temperature while Table
11.4 lists molar specific heat capacities of some gases.

From Table 11.3 you can note that water has the highest specific heat capacity compared to other substances. For this reason water is
used as a coolant in automobile radiators as well as a heater in hot water bags.

Owing to its high specific heat capacity, the water warms up much more slowly than the land during summer and consequently wind from the sea has a cooling effect. Now, you can tell why in desert areas, the earth surface warms up quickly during the day and cools quickly at night.

bbul" CALORIMETRY"

A system is said to be isolated if no exchange or transfer of heat occurs between the system and its surroundings. When different parts of an isolated system are at different temperature, a quantity of heat transfers from the part at higher temperature to the part at lower temperature.

The heat lost by the part at higher temperature is equal to the heat gained by the part at lower temperature. Calorimetry means measurement of heat.

When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, provided no heat is allowed to escape to the surroundings. A device in which heat measurement can be made is called a "calorimeter."

It consists a metallic vessel and stirrer of the same material like copper or alumiunium. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass wool etc.

The outer jacket acts as a heat shield and reduces the heat loss from the inner vessel. There is an opening in the outer jacket through which a mercury thermometer can be inserted into the calorimeter.

The following example provides a method by which the specific heat capacity of a given solid can be determinated by using the principle, heat gained is equal to the heat lost.
Q 3280512417

sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.
Class 11 Chapter 11 Example 3
Solution:

In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.
Mass of aluminium sphere (m_1) = 0.047 kg
Initial temp. of aluminium sphere = 100 °C
Final temp. = 23 °C
Change in temp (ΔT ) = (100 °C - 23 °C) = 77 °C
Let specific heat capacity of aluminium be s_(Al).
The amount of heat lost by the aluminium
sphere = m_1S_(Al) DeltaT = 0.047kg xxS_(Ai)xx77^oC
Mass of water (m_2) = 0.25 kg
Mass of calorimeter (m_3) = 0.14 kg
Initial temp. of water and calorimeter = 20 °C
Final temp. of the mixture = 23 °C
Change in temp. (ΔT_2) = 23 °C – 20 °C = 3 °C
Specific heat capacity of water (s_w)
= 4.18 \ \10^3 J kg^(–1) K^(–1)
Specific heat capacity of copper calorimeter
= 0.386 \ \ 10^3 J kg^(–1) K^(–1)
The amount of heat gained by water and calorimeter = m_2 sw ΔT_2 + m_3s_(cu_ΔT_2
= (m_2_(sw) + m)3s_(cu)) (ΔT_2)
= 0.25 kg \ \4.18 \ \10^3 J kg^(–1) K^(–1) + 0.14 kg
0.386 \ \ 10^3 J kg^(–1) K^(–1)) (23 °C – 20 °C)
In the steady state heat lost by the aluminium
sphere = heat gained by water + heat gained by calorimeter.
So, 0.047 kg s_(Al) 77 °C
= (0.25 kg 4.18 10^3 J kg^(–1) K^(–1)+ 0.14 kg
0.386 10^3 J kg^(–1) K^(–1))(3 °C)
s_(Al) = 0.911 kJ kg ^(–1) K^(–1)

### CHANGE OF STATE

Matter normally exists in three states: solid, liquid, and gas. A transition from one of these states to another is called a "change of state."

Two common changes of states are solid to liquid and liquid to gas (and vice versa). These changes can occur when the exchange of heat takes place between the substance and its surroundings.

Let us perform the following activity Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice.

Draw a graph between temperature and time (Fig. 11.9). You will observe no change in the temperature so long as there is ice in the beaker. In the above process, the temperature of the system does not change even though heat is being continuously supplied. The heat supplied is being utilised in changing the state from solid (ice) to liquid (water).

The change of state from solid to liquid is called melting and from liquid to solid is called fusion. It is observed that the temperature remains constant until the entire amount of the solid substance melts.

That is, both the solid and liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid. The temperature at which the solid and the liquid states of the substance in thermal equilibrium with each other is called its melting point.

It is characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atomspheric pressure is called its normal melting point. Let us do the following activity to understand the process of melting of ice.

Take a slab of ice. Take a metallic wire and fix two blocks, say 5 kg each, at its ends. Put the wire over the slab as shown in Fig. 11.10.

You will observe that the wire passes through the ice slab. This happens due to the fact that just below the wire, ice melts at lower temperature due to increase in pressure. When the wire has passed, water above the wire freezes again.

Thus the wire passes through the slab and the slab does not split. This phenomenon of refreezing is called regelation. Skating is possible on snow due to the formation of water below the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

100 °C when it again becomes steady. The heat supplied is now being utilised to change water from liquid state to vapour or gaseous state.

The change of state from liquid to vapour (or gas) is called vaporisation. It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour.

That is, both the liquid and vapour states of the substance coexist in thermal equilibrium, during the change of state from liquid to vapour. The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point.

Let us do the following activity to understand the process of boiling of water.
Take a round-bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask (Fig. 11.11).

As water gets heated in the flask, note first that the air, which was dissolved in the water, will come out as small bubbles. Later, bubbles of steam will form at the bottom but as they rise to the cooler water near the top, they condense and disappear.

Finally, as the temperature of the entire mass of the water reaches 100 °C, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible but as it comes out of the flask, it condenses as tiny droplets of water, giving a foggy appearance.

If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling begins again.

Thus boiling point increases with increase in pressure Let us now remove the burner. Allow water to cool to about 80 °C. Remove the thermometer and steam outlet. Close the flask with the airtight cork.

Keep the flask turned upside down on the stand. Pour ice-cold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure.

This explains why cooking is difficult on hills. At high altitudes, atmospheric pressure is lower, reducing the boiling point of water as compared to that at sea level.

On the other hand, boiling point is increased inside a pressure cooker by increasing the pressure. Hence cooking is faster. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point.

However, all substances do not pass through the three states: solid-liquid-gas. There are certain substances which normally pass from the solid to the vapour state directly and vice versa.

The change from solid state to vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime. Dry ice (solid CO_2) sublimes, so also iodine. During the sublimation process both the solid and vapour states of a substance coexist in thermal equilibrium.

### Latent Heat

As we have learnt that certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state.

The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process. For example, if heat is added to a given quantity of ice at –10 °C, the temperature of ice increases until it reaches its melting point (0 °C).

At this temperature, the addition of more heat does not increase the temperature but causes the ice to melt, or changes its state. Once the entire ice melts, adding more heat will cause the temperature of the water to rise.

A similar situation occurs during liquid gas change of state at the boiling point. Adding more heat to boiling water causes vaporisation, without increase in temperature.

The heat required during a change of state depends upon the heat of transformation and the mass of the substance undergoing a change of state. Thus, if mass m of a substance undergoes a change from one state to the other, then the quantity of heat required is given by

color{blue}(Q=mL)
orcolor{orange}(L =Q //m) .....(11.13)

where L is known as latent heat and is a characteristic of the substance. Its SI unit is J kg^(–1).

The value of L also depends on the pressure. Its value is usually quoted at standard atmospheric pressure. The latent heat for a solidliquid state change is called the latent heat of fusion (L_f), and that for a liquid-gas state change is called the "latent heat" of vaporisation (L_v).

These are often referred to as the heat of fusion and the heat of vaporisation. A plot of temperature versus heat energy for a quantity of water is shown in Fig. 11.12. The latent heats of some substances, their freezing and boiling points, are given in Table 11.5.

Table 11.5

"Note" that when heat is added (or removed) during a change of state, the temperature remains constant. Note in Fig. 11.12 that the slopes of the phase lines are not all the same, which indicates that specific heats of the various states are not equal.

For water, the latent heat of fusion and vaporisation are L_f = 3.33xx 10^5 J kg^(–1) and L_v = 22.6xx 10^5 J kg^(–1) respectively.

That is 3.33 105 J of heat are needed to melt 1 kg of ice at 0 °C, and 22.6 10^5 J of heat are needed to convert 1 kg of water to steam at 100 °C. So, steam at 100 °C carries 22.6xx 10^5 J kg^(–1) more heat than water at 100 °C. This is why burns from steam are usually more serious than those from boiling water.
Q 3210612519

When 0.15 kg of ice of 0 °C mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Calculate the heat of fusion of ice. (s_("water") = 4186 J kg^(–1) K^(–1))
Class 11 Chapter 11 Example 4
Solution:

Heat lost by water = ms_w (θ_f–θ_i)_w
= (0.30 kg ) (4186 J kg^(–1) K^(–1)) (50.0 °C – 6.7 °C)
= 54376.14 J
Heat required to melt ice = m_2L_f = (0.15 kg) L_f Heat required to raise temperature of ice water to final temperature = m_Is_w (θ_f–θ_i)_I
= (0.15 kg) (4186 J kg^(–1) K ^(–1)) (6.7 °C – 0 °C)
= 4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 kg) L_f + 4206.93 J
L_f = 3.34105 J kg^(–1).
Q 3270712616

Calculate the heat required to convert 3 kg of ice at –12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^(–1) K^(–1), specific heat capacity of water = 4186 J kg^(– 1) K^(–1), latent heat of fusion of ice = 3.35 105 J kg^(–1) and latent heat of steam = 2.256 106 J kg^(–1.)
Class 11 Chapter 11 Example 5
Solution:

We have
Mass of the ice, m = 3 kg
specific heat capacity of ice, s_(ice)
= 2100 J kg^(–1) K^(–1)
specific heat capacity of water, s_("water")
= 4186 J kg^(–1) K^(–1)
latent heat of fusion of ice, L_(f ice)
= 3.35 10^5 J kg^(–1)
latent heat of steam, L_("steam")
2.256 10^6 J kg^(–1)

Now, Q = heat required to convert 3 kg of
ice at –12 °C to steam at 100 °C,
Q_1 = heat required to convert ice at –12 °C to ice at 0 °C.
= m s_(ice) ΔT_1 = (3 kg) (2100 J kg^–1.
K^(–1)) [0–(–12)]°C = 75600 J
Q_2 = heat required to melt ice at
0 °C to water at 0 °C
= 1005000 J
Q_3 = heat required to convert water
at 0 °C to water at 100 °C.
= msw ΔT_2 = (3kg) (4186J kg^(–1) K^(–1))
(100 °C)
= 1255800 J
Q_4 = heat required to convert water
at 100 °C to steam at 100 °C.
= mL_("steam") = (3 kg ) (2.25610^6)J kg–1)
= 6768000 J
So, Q = Q_1 + Q_2 + Q_3 + Q_4
= 75600J + 1005000 J
+ 1255800 J + 6768000 J
= 9.1106 J