CBSE Class 8 Science Heat Chapter Notes

This Class 8 Science chapter notes page on Heat covers the difference between heat and temperature, the Celsius and Kelvin temperature scales with their conversion formulas, the three modes of heat transfer (conduction, convection and radiation), thermodynamic processes, units of heat, heat capacity and specific heat capacity, the consequences of water's high specific heat capacity, latent heat of fusion and vaporisation, and the solved exercises on heat with answers.

HEAT

Heat is a form of energy. It makes a substance hotter. Heat is measured by the temperature effect which it produces. When heat is given to a substance, its temperature increases (it becomes hotter), and when heat is removed from a substance, its temperature decreases (it becomes cold). When we supply heat energy to a substance, the kinetic energy of its molecules increases and they move with greater speeds at higher temperatures. On the other hand, if we remove heat energy from a substance, the kinetic energy of its molecules decreases and they move with lesser speeds at lower temperatures. Thus, we can consider heat to be the energy of molecular motion. Heat flows from a hotter substance to a colder substance. Heat energy is measured in the units of joules (J).

TEMPERATURE

Temperature is the degree of hotness (or coldness) of a substance. A hot substance is said to have a high temperature, whereas a cold substance is said to have a low temperature. The temperature of a substance is an indicator of the average kinetic energy of the molecules of the substance. In other words, temperature is a measure of heat energy level. The temperature of a body governs the flow of heat. Heat always flows from a body at higher temperature to a body at a lower temperature, and it keeps on flowing until both the bodies attain the same temperature. Temperature is measured in the units of degrees Celsius (°C) or kelvin (K).

We usually get confused between the terms 'heat' and 'temperature'. Let us take an example to understand their difference. Suppose we have a vessel of boiling water and a red-hot spark from a fire. The boiling water contains much more heat energy than the spark from a fire, but its temperature is much less than the spark of fire. The high temperature spark from a fire will not burn our hand much (because it contains much less heat energy), but the boiling water will burn our hand badly (because it contains much more heat energy).

[Figure: A rising curve of energy against temperature, starting from the label '0 K (absolute zero)', with the labels 'Rising energy' and 'Rising temperature'. Caption: At absolute zero, an object has no heat energy. But, above absolute zero, its molecules start to vibrate and have heat energy.]

[Photograph: A forest fire. Caption: A forest fire shows the power of uncontrolled heat.]

DIFFERENCES BETWEEN HEAT AND TEMPERATURE

TEMPERATURE SCALES

Just as two fixed marks (of 0 cm and 100 cm) are required on a metre scale for the measurement of length, two fixed temperatures are required for a temperature scale for the measurement of temperature. These two fixed temperatures are called the 'fixed points' of the temperature scale. The two fixed temperatures (or fixed points) which can be easily reproduced are:
i) The temperature at which pure ice melts under standard atmospheric pressure is taken as the lower fixed point of the temperature scale. In other words, the melting point of pure ice is taken as the lower fixed temperature.
ii) The temperature at which pure water boils under standard atmospheric pressure is taken as the upper fixed point of the temperature scale. In other words, the boiling point of pure water is taken as the upper fixed temperature.

When water boils, steam is formed. The boiling water and the steam formed from it are at the same temperature. So, the temperature of steam formed from water boiling under standard atmospheric pressure is also taken as the upper fixed point of the temperature scale.

It is necessary to write 'standard atmospheric pressure' in the definitions of the lower fixed point and the upper fixed point, because both the melting point of ice and the boiling point of water vary with the change in atmospheric pressure.

The distance between the lower fixed point and the upper fixed point is called the fundamental interval. The fundamental interval is divided into a number of equal divisions. Each division is called 'one degree'. There are two temperature scales which are commonly used in the measurement of temperature these days:
i) Celsius scale of temperature, and
ii) Kelvin scale of temperature.

The Kelvin temperature scale is also called the absolute temperature scale.

CELSIUS SCALE OF TEMPERATURE

The Celsius scale is a convenient, metric scale of temperature which was given by Celsius. On the Celsius scale, the melting point of ice is given the value of 0° (zero degree), and the temperature of boiling water (or steam) is given the value of 100° (hundred degrees). In other words, on the Celsius scale the lower fixed point is 0°C and the upper fixed point is 100°C, so there are 100 - 0 = 100 equal divisions, or 100 degrees, between the two fixed temperatures. So, in this case, the fundamental interval is divided into 100 equal divisions (or degrees). The two fixed temperatures of the Celsius scale have been given simple values of 0 and 100, and there are 100 degrees in it, so it is quite easy to do calculations using Celsius temperatures. For convenience, the temperatures in laboratories and in everyday life are measured in 'degrees Celsius' (°C).

[Figure: A Celsius thermometer with reference temperatures marked in kelvin and degrees Celsius.]

If a Celsius thermometer is needed to measure temperatures below the lower fixed point (below 0°C), the scale markings are continued below the 0°C mark. Temperatures below 0°C are written with a minus sign (like -10°C or -20°C, etc.). Similarly, to measure temperatures higher than the upper fixed point (above 100°C), the scale markings are continued above the 100°C mark. The distance between consecutive marks on the Celsius scale remains the same.

KELVIN SCALE OF TEMPERATURE

The temperatures on the Celsius scale can be positive (above 0°C) as well as negative (below 0°C). On the Kelvin scale, all the temperatures are always positive. There can be no minus temperatures on the Kelvin scale.

It has been found that the lowest temperature which can be attained in a laboratory is minus 273°C (-273°C). This lowest possible temperature is called 'absolute zero'. Thus, the temperature of -273°C (which is the lowest temperature that can be reached) is taken as the 0 (zero) of the Kelvin scale. The size of each division on the Kelvin scale is the same as that on the Celsius scale, so:
-273°C = 0 K
And hence, 0°C = 273 K

Thus, a temperature of 0° on the Celsius scale is equal to 273 on the Kelvin scale. The word 'degree' or its sign (°) is not written while writing Kelvin temperatures. We just write K for kelvin. For example, we write 273 K, and not 273°K.

[Figure: Two thermometers side by side. The Celsius thermometer is marked 0°C, 37°, 50° and 100°. The Kelvin thermometer is marked 273 K, 300, 350 and 373.]

Temperatures measured on the Celsius scale can be converted into Kelvin temperatures by adding the figure 273 to each reading on the Celsius scale. For example:
100°C = (100 + 273) K = 373 K

The lower fixed point on the Kelvin scale is the temperature of pure melting ice under standard atmospheric pressure, and its value is 273 K. This means that the melting point of ice has a value of 273 K on the Kelvin scale of temperature.

The upper fixed point on the Kelvin scale is the temperature of boiling water (or steam) under standard atmospheric pressure, and its value is 373 K. This means that the boiling point of water (or steam) has a value of 373 K on the Kelvin scale.

A comparison between the Celsius and Kelvin scales of temperature is shown in the figure.

[Figure: Comparison of the Celsius and Kelvin scales. (a) Celsius scale and (b) Kelvin scale. Upper fixed point (boiling point of water): 100°C and 373 K. Lower fixed point (melting point of ice): 0°C and 273 K. Lowest attainable temperature: -273°C and 0 K. Caption: Comparison between Celsius and Kelvin scales of temperature.]

It is clear from the figure that:
-273°C = 0 K
0°C = 273 K
and 100°C = 373 K

One advantage of the Kelvin scale over the Celsius scale is that all the temperatures on the Kelvin scale are positive. Kelvin is the SI unit of temperature and is denoted by the letter K. Since the fundamental intervals of the Celsius scale and the Kelvin scale have each been divided into 100 divisions, 1 degree Celsius is equal to 1 division of the Kelvin scale. These days the Kelvin scale of temperature is used for all scientific work.

RELATION BETWEEN CELSIUS SCALE AND KELVIN SCALE

The relation between the Kelvin scale and the Celsius scale can be written as:
Temperature on Kelvin scale = Temperature on Celsius scale + 273
or K = C + 273
where K = Temperature on Kelvin scale
and C = Temperature on Celsius scale

This relation can be used to convert a Celsius temperature into a Kelvin temperature, or a Kelvin temperature into a Celsius temperature.

Conversion of Celsius to Fahrenheit scale: On the Fahrenheit scale, the ice-point and the steam-point are called 32°F and 212°F, and their interval has been divided into 180 degrees. The following relation holds for the conversion of the Celsius scale to the Fahrenheit scale:
C / 100 = (F - 32) / 180
or C / 5 = (F - 32) / 9

TRANSFER OF HEAT

To carry heat from one part of an object to its other part, or from one object to another object, is called transfer of heat. Heat can be transferred from a hot object to a cold object in three different ways:
i) by conduction
ii) by convection, and
iii) by radiation

Conduction: Conduction is the transfer of heat from the hotter part of a material to its colder part (or from a hot material to a cold material in contact with it) without the movement of the material as a whole. In all solids, heat is transferred by the process of conduction.
Examples:
i) A cold metal spoon dipped in a hot cup of tea gets heated by conduction.
ii) A frying pan kept on a gas stove transfers the heat of the gas through its metal bottom by the process of conduction.

Convection: Convection is the transfer of heat from the hotter parts of a liquid (or gas) to its colder parts by the movement of the liquid (or gas) itself. The transfer of heat by convection can take place only in liquids and gases, and not in solids, because the particles in liquids and gases can move about freely.
Examples:
i) A room heater heats all the air in a room by setting up convection currents in the air.
ii) Water is a poor conductor of heat, so it cannot transfer heat by conduction. Water transfers heat by the process of convection and gets heated up.

Radiation: Radiation is the transfer of heat energy from a hot body to a cold body by means of heat rays, without any material medium between them.
Examples:
i) The sun's heat reaches the earth by the process of radiation.
ii) When we stand next to a burning fire, we can feel the heat of the fire. This heat, transferred from the fire to our face, is by the process of radiation.

THERMODYNAMIC PROCESS

The operation by which a thermodynamic system changes from one state to another is called a thermodynamic process.
i) Isothermal process: A process in which, although heat enters or leaves the system, the temperature of the system remains constant throughout the process is called an isothermal process. For an isothermal process, change in temperature (ΔT) = 0. Changes of state (e.g. freezing, melting, evaporation and condensation) are all examples of isothermal processes.
ii) Adiabatic process: A process during which no heat enters or leaves the system during any step of the process is known as an adiabatic process. A reaction carried out in an isolated system is an example of an adiabatic process. For an adiabatic process, change in heat (q) = 0, or q remains constant.
iii) Isobaric process: A process during which the pressure of the system remains constant throughout the reaction is called an isobaric process. For example, heating of water to its boiling point, and its vaporisation taking place at the same atmospheric pressure. Expansion of a gas in an open system is an example of an isobaric process. For an isobaric process, change in pressure (ΔP) = 0.
iv) Isochoric process: A process during which the volume of the system remains constant throughout the reaction is known as an isochoric process. The heating of a substance in a non-expanding chamber, or any change taking place in a closed system, are examples of isochoric processes. For an isochoric process, change in volume (ΔV) = 0.

UNITS OF HEAT

Like other forms of energy, the S.I. unit of heat is the joule (J). The other most commonly used unit of heat is the calorie (symbol: cal). It is defined as follows:
One calorie of heat is the quantity of heat required to raise the temperature of 1 g of water through 1°C.

In this definition of the calorie, it has been assumed that the heat required to raise 1 g of water through 1°C at any temperature is the same. However, this is not true. The reason is that the thermal expansion of water is not uniform near 4°C. Water shows uniform and smooth expansion only beyond 14°C. Hence the correct definition of the calorie is given as follows:
One calorie of heat is the quantity of heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.

The unit calorie is related to the S.I. unit joule as follows:
1 calorie (or 1 cal) = 4.1855 J or 4.2 J

Sometimes calorie is a smaller unit of heat, so we use another unit called the kilo-calorie. One kilo-calorie of heat is the quantity of heat required to raise the temperature of 1 kg of water from 14.5°C to 15.5°C.
Thus, 1 kilo-calorie = 1000 calorie
and 1 kilo-calorie = 4185.5 J or 4200 J

The unit kilo-calorie is generally used for measuring the energy value of foods.

HEAT CAPACITY

From our everyday experience we know that different bodies require different amounts of heat to raise their temperature by the same interval. For example, if equal masses of water and copper are heated through 1°C, the amount of heat required for water is more than that required for copper. This property of a substance is expressed in terms of heat capacity. In other words, bodies of different materials, but of the same mass, have different heat capacities.

The heat capacity of a body is the amount of heat required to raise its temperature by 1°C.

Unit of heat capacity: By definition,
Heat capacity = Amount of heat / Rise in temperature
Thus, the S.I. unit of heat capacity is joule per degree Celsius (i.e. J °C^-1).

The other unit is calorie °C^-1 or kilo-calorie °C^-1.

SPECIFIC HEAT CAPACITY

The heat capacity of a body when expressed for unit mass is called its specific heat capacity. Thus, the specific heat capacity of a substance is defined as follows:
The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of that substance through 1°C.

It is denoted by the symbol c.

Unit of specific heat capacity: By definition,
Specific heat capacity c = Amount of heat / (Mass × Rise in temperature)
Thus, the S.I. unit of specific heat capacity is joule per kilogram per degree Celsius (J kg^-1 °C^-1 or J kg^-1 K^-1). The other unit of specific heat capacity is cal g^-1 °C^-1 or kilo-cal kg^-1 °C^-1. These units are related as:
1 cal g^-1 °C^-1 = 1 kilo-cal kg^-1 °C^-1 = 4.2 × 10³ J kg^-1 °C^-1

The specific heat capacity of a substance is its characteristic property: it is different for different substances. If we heat equal masses of two different substances on the same burner, so that the rate of heat supply is the same, we notice that in the same time interval the rise in temperature for the two substances is different. This is clearly due to their different specific heat capacities. Water has an unusually high specific heat capacity (= 4200 J kg^-1 °C^-1).

The table below gives the specific heat capacity of some substances.

RELATIONSHIP BETWEEN HEAT CAPACITY AND SPECIFIC HEAT CAPACITY

If Q joule of heat energy is supplied to M kg of a substance so that its temperature rises through Δt °C, then its specific heat capacity c is given as:
c = Q / (M × Δt) J kg^-1 °C^-1

Thus, the amount of heat Q = Mc Δt joule
= mass × specific heat × change in temperature

The quantity of heat required to change the temperature of a body by 1°C is called the heat capacity of the body.
Heat capacity = Q / Δt = Mc
= mass × specific heat capacity

CONSEQUENCES OF HIGH SPECIFIC HEAT CAPACITY OF WATER

Some consequences of the high specific heat capacity of water are given below:
i) The specific heat capacity of water is very high, equal to 1000 cal kg^-1 °C^-1 or 4200 J kg^-1 °C^-1. It is about five times as high as that of sand. Hence, the amount of heat required by a certain mass of water is nearly five times more than that required by the same mass of sand to raise it by the same temperature interval. Similarly, a certain mass of water will give out nearly five times more heat than sand of the same mass for the same fall in temperature. As such, sand or earth is heated or cooled much more quickly than water. Thus there is a large difference in temperature between the land and the sea, due to which land and sea breezes are formed. It makes the climate near the sea-shore moderate.
ii) Hot water bottles are used for fomentation, since water does not cool quickly due to its large specific heat capacity.
iii) Water is used as an effective coolant by flowing it in pipes around the heated parts of machines (e.g. the radiator in a car and the generator are filled with water). This is because, due to its large specific heat capacity, it can extract more heat.
iv) In cold countries, wine and juice bottles are placed under water to avoid their freezing, because water does not cool quickly owing to its high specific heat capacity.
v) Farmers fill their fields with water to protect the crops from frost. Water, having a high specific heat capacity, does not allow the temperature in the surrounding area of the plants to fall below 0°C. In the absence of water, if on a cold night the temperature falls below 0°C, the water in the fine capillaries of plants will freeze and the veins will burst due to the increase in volume of water on freezing. As a result, the plants will die and the crop will be destroyed.

LATENT HEAT

During the change of state of a substance without change in its temperature, a considerable amount of heat is absorbed or liberated. Heat is absorbed by a substance during melting, and an equal amount of heat is liberated by the substance during freezing, without showing any rise or fall in temperature. Similarly, heat is absorbed by a liquid during vaporisation, and an equal amount of heat is liberated by the vapour during condensation, without showing any rise or fall in temperature.

Since this heat energy is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.

The latent heat, when expressed for unit mass of the substance, is called the specific latent heat. Specific latent heat is denoted by the symbol L.
Specific latent heat L = Heat exchanged for change of state / mass
or L = Q / M
Q = mass m × specific latent heat L

In other words, L is the amount of heat absorbed or liberated per unit mass for the change of state. The latent heat is used up in increasing the potential energy of the molecules of the substance.

Explanation: According to the kinetic theory, the molecules in a solid vibrate about their mean positions. The total energy of a molecule is the sum of its kinetic energy (which depends on temperature) and its potential energy (which depends on the force of attraction between the molecules and the distance between them). When a solid changes into a liquid without change in temperature, the average kinetic energy of the molecules does not change, but the potential energy increases, as the distance between the molecules on an average increases. The energy is required for increasing the separation, as work is done against the attractive force between the molecules. The latent heat of fusion is thus utilised in increasing the potential energy of the molecules.

Similarly, during the change of state from liquid to vapour without change in temperature, the average kinetic energy of the molecules does not change. Since the vapour occupies a much larger volume than the liquid (i.e. the intermolecular separation increases), a part of the heat energy (latent heat of vaporisation) supplied is used up in increasing the potential energy of the molecules (i.e. in doing work against the force of attraction amongst the molecules to increase the intermolecular separation), and the rest of the energy is used up in doing work for the expansion (i.e. increase in volume) against the atmospheric pressure.

SPECIFIC LATENT HEAT OF FUSION

The specific latent heat of fusion of a substance is defined as the quantity of heat required to convert unit mass of the substance from the solid to the liquid state without change of temperature.

Unit: Its S.I. unit is J kg^-1. The other common unit is cal g^-1 or kilo-calorie kg^-1.

For ice, the specific latent heat of fusion is 336000 J kg^-1 (= 80 cal g^-1). It means that 1 kg of ice at 0°C absorbs 336000 J of heat energy to convert into water at 0°C, or 1 kg of water at 0°C will liberate 336000 J of heat to convert into ice at 0°C.

The table below gives the specific latent heat of fusion of some common substances.

SPECIFIC LATENT HEAT OF VAPORISATION

The specific latent heat of vaporisation of a substance is defined as the quantity of heat required to convert unit mass of the substance from the liquid to the vapour state without change of temperature.

Unit: Its S.I. unit is J kg^-1. The other common unit is cal g^-1 or kilo-cal kg^-1.

The specific latent heat of vaporisation of steam is 2260000 J kg^-1 (= 540 cal g^-1). This means that 2260000 J of heat is needed to convert 1 kg of water at 100°C to steam at 100°C, or 2260000 J of heat is to be withdrawn to convert 1 kg of steam at 100°C to water at 100°C.

The figure below illustrates the changes in the state of 1 g of ice, initially at -10°C, when it is heated at a constant rate.

[Figure: Temperature (°C) against TIME (s) for 1 g of ice initially at -10°C, with the flat parts labelled MELTING and VAPORISATION. Part AB shows the continuous rise in temperature up to the melting point 0°C at B, with heat absorbed 21 J. Part BC shows melting of ice at constant temperature 0°C, complete at C, with heat absorbed 336 J. Part CD shows the continuous rise in temperature of water up to the boiling point 100°C, with heat absorbed 420 J. Part DE shows vaporisation at constant temperature 100°C, complete at E, with heat absorbed 2260 J. The part beyond E shows the rise in temperature of steam.]

Figure: Change of state of ice from -10°C

The melting of ice occurs at 0°C and the boiling of water occurs at 100°C. The part DE is longer than the part BC, because the specific latent heat of vaporisation is much more than the specific latent heat of fusion. If the ice is initially at 0°C, the origin of the graph would be at B.

Natural consequences of high specific latent heat of fusion of ice: Ice has a high specific latent heat of fusion (i.e. 336000 J kg^-1). It is due to this fact that snow on the mountains does not melt all at once, but changes into water slowly as it gets heat from the sun. If the latent heat were not so high, all the snow would melt very quickly even with a small amount of heat, and there would be floods in rivers. Similarly, water in lakes and ponds in cold countries does not freeze all at once. It freezes slowly and thus keeps the surroundings moderate.

Drinks are cooled by adding pieces of ice rather than ice-cooled water at 0°C. This is because 1 g of ice takes 336 J of heat from the drink to melt into water at 0°C. Thus the drink loses a considerable amount of heat and can be cooled effectively even with a small quantity of ice.

Consequences of high specific latent heat of steam: Steam causes more severe burns than boiling water, both being at the same temperature (i.e. 100°C). This is because the steam first condenses into water at 100°C by imparting its latent heat (= 2260 J g^-1), and then its effect is the same as that of boiling water. Obviously, 1 g of steam at 100°C imparts 2260 J of extra heat compared with 1 g of water at 100°C.

Because of its high specific latent heat, steam is used for running trains or machines, where the heat released during condensation of steam is converted to mechanical energy.

EXERCISES

Answer the following questions:

1. An iron ball requires 5000 J heat to raise its temperature by 10°C. Calculate the heat capacity of the iron ball.

Answer:
1. Heat supplied Q = 5000 J and rise in temperature Δt = 10°C.
2. Heat capacity = Q / Δt = 5000 / 10 = 500 J °C^-1.
3. Hence, the heat capacity of the iron ball is 500 J °C^-1.

Teacher's Note:
a) Heat capacity is the heat needed for a 1°C rise, so divide the heat by the rise in temperature.
b) The answer has no mass term, because heat capacity is for the whole body, not for unit mass.

2. Calculate the amount of heat required to raise the temperature of 2 kg from 10°C to 50°C. Specific heat capacity of water is 4200 J kg^-1 °C^-1.

Answer:
1. Mass m = 2 kg, specific heat capacity c = 4200 J kg^-1 °C^-1, and rise in temperature Δt = 50 - 10 = 40°C.
2. Heat Q = m × c × Δt = 2 × 4200 × 40 = 336000 J.
3. Hence, the heat required is 336000 J (or 3.36 × 10⁵ J).

Teacher's Note:
a) Find the rise in temperature first: 50°C minus 10°C gives 40°C.
b) Multiply mass, specific heat and temperature rise together.

3. A body of mass 500 g requires 3000 J of heat in order to raise its temperature from 35°C to 45°C. Calculate the specific heat capacity of the body.

Answer:
1. Mass m = 500 g = 0.5 kg, heat Q = 3000 J, and rise in temperature Δt = 45 - 35 = 10°C.
2. Specific heat capacity c = Q / (m × Δt) = 3000 / (0.5 × 10) = 3000 / 5 = 600 J kg^-1 °C^-1.
3. Hence, the specific heat capacity of the body is 600 J kg^-1 °C^-1.

Teacher's Note:
a) The mass must be in kilograms before using the S.I. unit J kg^-1 °C^-1, so 500 g becomes 0.5 kg.
b) The rise in temperature is 10°C, not 45°C.

4. Some heat is given to 120 g of water and its temperature rises by 10 K. When the same amount of heat is given to 60 g of an oil, its temperature rises by 40 K. The specific heat capacity of water is 4200 J kg^-1 K^-1. Calculate (i) the amount of heat in joule to water, (ii) the specific heat capacity of the oil.

Answer:
1. Mass of water m = 120 g = 0.12 kg, rise in temperature = 10 K, and c = 4200 J kg^-1 K^-1.
2. (i) Heat to water Q = m × c × Δt = 0.12 × 4200 × 10 = 5040 J.
3. The same heat is given to the oil, so Q = 5040 J.
4. For the oil, mass m = 60 g = 0.06 kg and rise in temperature Δt = 40 K.
5. (ii) Specific heat capacity of oil c = Q / (m × Δt) = 5040 / (0.06 × 40) = 5040 / 2.4 = 2100 J kg^-1 K^-1.
6. Hence, (i) the heat given to water is 5040 J, and (ii) the specific heat capacity of the oil is 2100 J kg^-1 K^-1.

Teacher's Note:
a) Convert both masses from grams to kilograms before calculating.
b) A rise of 1 K is the same as a rise of 1°C, so the same formula applies.
c) The oil's specific heat capacity comes out lower than water's, which matches the table of values.

5. An electric heater of power 1000 W raises the temperature of 5 kg of a liquid from 25°C to 31°C in 2 minutes. Calculate (i) heat capacity of the liquid and (ii) its specific heat capacity.

Answer:
1. Power P = 1000 W and time t = 2 minutes = 120 s.
2. Heat supplied Q = P × t = 1000 × 120 = 120000 J.
3. Rise in temperature Δt = 31 - 25 = 6°C.
4. (i) Heat capacity = Q / Δt = 120000 / 6 = 20000 J °C^-1.
5. (ii) Specific heat capacity c = Q / (m × Δt) = 120000 / (5 × 6) = 120000 / 30 = 4000 J kg^-1 °C^-1.
6. Hence, (i) the heat capacity of the liquid is 20000 J °C^-1, and (ii) its specific heat capacity is 4000 J kg^-1 °C^-1.

Teacher's Note:
a) Convert the time from minutes to seconds, because power is in watts (joules per second).
b) Heat capacity uses the total heat and temperature rise; specific heat capacity also divides by the mass.

6. How much heat is required to melt 5 kg of ice? Sp. latent heat of ice = 336 J g^-1.

Answer:
1. Mass m = 5 kg = 5000 g.
2. Heat required Q = m × L = 5000 × 336 = 1680000 J.
3. Hence, the heat required is 1680000 J (or 1.68 × 10⁶ J).

Teacher's Note:
a) The latent heat is given per gram, so the mass must first be converted from kilograms to grams.
b) This value is the same as 336000 J kg^-1 given in the chapter.

7. A 30 g ice cube at 0°C is dropped into 200 g of water at 30°C. Calculate the final temperature of water when all the ice cube has melted. Given: latent heat of ice = 80 cal/g; sp. heat capacity of water = 1 cal g^-1 °C^-1.

Answer:
1. Heat needed to melt the ice = m × L = 30 × 80 = 2400 cal.
2. Let the final temperature be T °C. Heat lost by the water in cooling from 30°C to T°C = 200 × 1 × (30 - T) = 6000 - 200T cal.
3. Heat gained by the melted ice (now water at 0°C) in warming to T°C = 30 × 1 × (T - 0) = 30T cal.
4. Heat gained by the ice must equal the heat lost by the water, so 2400 + 30T = 6000 - 200T.
5. Therefore, 230T = 3600, and T = 3600 / 230 = 15.65°C (approximately).
6. Hence, the final temperature of the water is about 15.7°C.

Teacher's Note:
a) First check whether the water has enough heat to melt the whole ice cube, which it does because 6000 cal is more than 2400 cal.
b) The melted ice becomes water at 0°C, and then it warms up with the rest of the water.
c) The total mass after melting is 230 g, which is why 230 appears in the denominator.

8. How much boiling water (at 100°C) is needed to melt 2 kg of ice so that the mixture, which is all water, is at 0°C. Given: sp heat capacity of water = 4.2 J g^-1 °C^-1, latent heat of ice = 336 J/g.

Answer:
1. Heat needed to melt 2 kg (2000 g) of ice at 0°C = 2000 × 336 = 672000 J.
2. Heat released by 1 g of boiling water when it cools from 100°C to 0°C = 1 × 4.2 × 100 = 420 J.
3. Let the mass of boiling water be m g. Then m × 420 = 672000, so m = 1600 g.
4. Hence, 1600 g (1.6 kg) of boiling water is needed.

Teacher's Note:
a) The heat that the boiling water gives out must equal the heat needed to melt the ice, because the final mixture is at 0°C.
b) Use grams throughout, because the specific heat and latent heat are given per gram.

9. One kilogram of ice at -10°C is heated until the whole of it vaporises. How much heat is required? Specific latent heat of fusion of ice = 336 × 10³ J/kg; specific latent heat of steam = 2268 × 10³ J/kg; specific heat capacity of ice = 2.1 × 10³ J kg^-1 °C^-1. Specific heat capacity of water = 4.2 × 10³ J kg^-1 °C^-1.

Answer:
1. Heat to warm the ice from -10°C to 0°C = 1 × 2.1 × 10³ × 10 = 21000 J.
2. Heat to melt the ice at 0°C = 1 × 336 × 10³ = 336000 J.
3. Heat to warm the water from 0°C to 100°C = 1 × 4.2 × 10³ × 100 = 420000 J.
4. Heat to vaporise the water at 100°C = 1 × 2268 × 10³ = 2268000 J.
5. Total heat Q = 21000 + 336000 + 420000 + 2268000 = 3045000 J.
6. Hence, the total heat required is 3045000 J, or about 3.045 × 10⁶ J.

Teacher's Note:
a) The process has four stages: warming the ice, melting it, warming the water, and vaporising the water. Add the heat for each stage.
b) The question gives 2268 × 10³ J/kg for steam, while the chapter's text gives 2260000 J/kg. The question's value is used here, so the answer is 3045000 J. With 2260000 J/kg the total would be 3037000 J.

100° celsius = 100 + 273 kelvin

                   = 273 kelvin

100° celsius = 100 + 273 kelvin

                   = 273 kelvin