ICSE Class 8 Physics Chapter 03 Heat

Heat

You have learnt certain things about heat in your earlier classes. For example, you have learnt how temperature is measured. You have learnt that heat can cause (a) changes in temperature, (b) changes in state, and (c) changes in the size of solids and in the volume of liquids. You also have some idea about the ways in which heat is transferred. In this chapter, we will revise some of these ideas and discuss them in greater detail. We will also define some of the quantities involved in these processes.

Heat Is A Form Of Energy

You will recall that energy is defined as the capacity to do work. Let us see how heat satisfies this definition. In order to make a motor vehicle move, petrol is burnt inside the engine to produce heat. A set of complex processes then converts this heat to the work of moving the car. Thus, heat has the capacity to do work and is a form of energy. In this example, heat performs the work of moving the car, and as the car begins to move, this work changes to the kinetic energy of the car.

Unit Of Heat

Since heat is a form of energy, it is measured in joules (J), which is the SI unit of work and energy. However, traditionally, heat is measured in calories (cal). The relation between the two quantities is

\[1 \text{ calorie} = 4.18 \text{ joules} = 4.2 \text{ joules}\]

The quantity 4.2 joules per calorie is called the mechanical equivalent of heat, and has the symbol J. This is because, under ideal conditions, 1 cal of heat can be converted into 4.2 J of (mechanical) work.

Another unit of heat called the kilocalorie (kcal) or Calorie (Cal) is still widely used, mainly in measuring the energy values of food, and the energy spent in exercise and other physical activities. As the name suggests, the kilocalorie is 1000 times the calorie. When speaking of the Calories expended in any activity, such as weightlifting or cycling, we usually say Calories 'burnt'.

Teacher's Note

When we eat food, we measure its energy content in kilocalories or "Calories" - this is why food packages show calorie counts, which directly relate to the heat energy our bodies use for daily activities.

\[1 \text{ Calorie} = 4.2 \times 10^3 \text{ joules}\]

A man weighing 70 kg climbs a three-storey building. Each storey has 20 steps of height 15 cm each. How many Calories of energy does he burn?

Given, mass of the man = 70 kg.

Therefore, weight of the man = (70 × 9.8) N.

This is the force the man overcomes while climbing.

Total height of stairs = 3 × 20 × 0.15 m = 9 m.

Then work done by the man = F × d = 70 × 9.8 × 9 = 6174 J.

Energy burnt by him = \(\frac{6174}{4.2 \times 10^3}\) = 1.47 Calories.

Calorimetry

The study of the heat gained or lost during physical processes and chemical reactions is called calorimetry. Temperature is of central importance in any such study since whenever a body (or a material) loses or gains heat, its temperature changes. Besides, heat flows by itself from a higher temperature to a lower temperature. You must remember that the direction in which heat flows between two bodies in contact depends only on their temperature. Thus, if a small body at a higher temperature is brought into contact with a much larger body at a lower temperature, heat will flow from the smaller to the larger body. Again, if two bodies in contact are at the same temperature, no heat will flow between them, irrespective of how hot or cold they are. This brings us to two very important points related to the flow of heat between two bodies.

Heat flows between two bodies in contact only if they are at different temperatures.

Heat stops flowing when the two bodies in contact reach the same temperature, or are in thermal equilibrium (equilibrium means balance).

Amount Of Heat

There is a definite relation between the heat gained (or lost) by a body and the change in its temperature. The following activities will help you understand this. However, even simple experiments involving heat can be dangerous. Therefore, you must do these only in the presence of an adult.

Pour a measured quantity of water into a small vessel. Record its temperature. Heat the vessel over a small flame for, say, a minute (use a watch to note the time). Then measure the temperature of the water again and find the change in temperature. Repeat the procedure, but let the water be heated for double the time. If possible, do it again and heat the water for thrice as long as the first time. You will find that the change in temperature in the second and third cases is about twice and three times the change in the first case. It should be obvious that the heat supplied in the second case and that in the third case are twice and three times as much as the heat supplied in the first case. Hence, we can conclude that the rise in temperature of a system (a solid or liquid) is proportional to the heat received by it.

Next pour a measured quantity of water into one vessel and pour double the quantity of water into another identical vessel. Measure the temperature of the water. Place both the vessels in a water bath (a large vessel of water) and heat the water for some time. (This way both the vessels of water will receive the same amount of heat.) You will find that the change in the temperature of the water in the second vessel is about half that of the water in the first vessel. Since the heat supplied in the two cases is the same, we can conclude that the rise in the temperature of a system is inversely proportional to its mass.

Lastly, pour equal quantities of oil and water into two identical vessels and heat them in a water bath for the same time. You will find that the temperature of the oil rises faster than that of the water. Hence, the rise in the temperature of a substance depends on the nature of the substance.

Specific Heat Capacity

We can now summarise what we have learnt so far mathematically. Let Q be the heat gained by a solid or liquid of mass m. Then the rise in its temperature (θ) will be related to Q and m as follows.

\[Q = m\theta \times \text{(something related to the nature of the substance)}\]

This 'something related to the nature of the substance' is called the specific heat capacity or specific heat of the substance, and has the symbol s.

Now we can write the relation (1) as

\[Q = ms\theta\]

Then

\[s = \frac{Q}{m\theta}\]

If Q is in joules (J), m in kg and θ in °C, the unit of s is J/kg °C. The relation (3) also helps us define specific heat. The specific heat of a substance is the heat required to raise the temperature of 1 kg of the substance by 1°C. The relations (2) and (3) hold for a fall in temperature as well as for a rise in temperature. Hence, s is also the amount of heat 1 kg of a substance loses when its temperature falls by 1°C.

Teacher's Note

Water's high specific heat capacity is why coastal areas have milder climates - the ocean absorbs and releases heat slowly, moderating temperature changes throughout the year.

Different materials have different specific heats. Some of these are listed in the table. Water has an unusually high specific heat. This means it has to gain (or lose) a large amount of heat for a small rise (or fall in temperature). This makes water useful in the study of heat and in many other ways, some of which are mentioned here.

It is used as a coolant in car radiators, factories and power stations because it can absorb a lot of heat without becoming too hot or boiling.

A hot-water bottle remains hot for a long time because water has to lose a lot of heat in order to cool down.

The high specific heat of water causes land and sea breezes. The sea takes much longer to get heated and to cool down than the land (the specific heat of land is much lower than that of water). Hence, it is cooler than the land during the day, and warmer than the land during the night. This makes sea breezes blow in during the day and land breezes blow out during the night.

The temperature of a vessel weighing 250 g increases by 20°C when it absorbs 2000 J of heat. Find the specific heat capacity of the material of the vessel.

Here, m = 250 g = 0.25 kg, θ = 20°C, Q = 2000 J.

You know that \(s = \frac{Q}{m\theta} = \frac{2000 \text{ J}}{0.25 \text{ kg} \times 20°\text{C}} = 400 \text{ J/kg °C}\).

How much heat would be required to raise the temperature of half a litre of water from 25°C to 35°C. The specific heat capacity of water is 4200 J/kg°C.

Here, volume of water = 0.5 L = 500 cc = 500 × 10⁻⁶ m³ = 5 × 10⁻⁴ m³.

Therefore, mass of water = volume × density = 5 × 10⁻⁴ m³ × 1000 kg/m³ = 0.5 kg.

Heat required = ms\theta = 0.5 kg × (4200 J/kg°C) × 10°C = 21,000 J.

Find the mass of a glass bowl of specific heat capacity 600 J/kg°C if its temperature rises by 20°C when it absorbs 1200 J of heat.

\(m = \frac{Q}{s\theta} = \frac{1200 \text{ J}}{(600 \text{ J/kg°C}) \times 20°\text{C}} = 0.1 \text{ kg} = 100 \text{ g}\).

A metallic cylinder of mass 100 g absorbs 1500 J of heat. Find the rise in its temperature if the specific heat capacity of the metal is 500 J/kg°C.

\(\theta = \frac{Q}{ms} = \frac{1500 \text{ J}}{0.1 \text{ kg} \times 500 \text{ J/kg°C}} = 30°\text{C}\).

Heat Capacity

The heat capacity of a body is the amount of heat required to raise its temperature by 1°C. It is also the heat lost by the body when its temperature falls by 1°C. We know that the heat required to raise the temperature of a body of mass m by θ°C is

\[Q = ms\theta\].

If we make θ = 1°C in this equation,

\[\text{Heat capacity} = ms\]

It follows that

\[Q = \text{heat capacity} \times \theta\]

The unit of heat capacity is joules per degree celsius or J/°C.

Find the heat capacity of the vessel described in Example 2.

Heat capacity = \(ms = \frac{Q}{\theta} = \frac{2000 \text{ J}}{20°\text{C}} = 100 \text{ J/°C}\).

Find the heat capacity of a copper vessel of mass 120 g. Take the specific heat of copper as 400 J/kg°C.

Heat capacity = \(ms = (120 \times 10^{-3}) \text{ kg} \times 400 \text{ J/kg°C} = 48 \text{ J/°C}\).

A metal ball of heat capacity 50 J/°C loses 2000 J of heat. By how much will its temperature fall?

Heat capacity = \(\frac{Q}{\theta}\)

Therefore, \(\theta = \frac{Q}{\text{heat capacity}} = \frac{2000 \text{ J}}{50 \text{ J/°C}} = 40°\text{C}\).

A vessel of milk of heat capacity 200 J/°C was kept in a refrigerator for three hours. If its temperature fell by 15°C, how much heat did it lose per hour?

Given, heat capacity = 200 J/°C, θ = 15°C.

Therefore, \(Q = (200 \text{ J/°C}) \times 15°\text{C} = 3000 \text{ J}\).

This heat is lost in three hours.

Therefore, heat lost per hour = \(\frac{3000 \text{ J}}{3} = 1000 \text{ J}\).

Teacher's Note

Understanding heat capacity helps explain why a full bathtub takes much longer to heat or cool than a cup of water - larger masses have greater heat capacities.

This is a preview of the first 3 pages. To get the complete book, click below.