5. Heat
Complete solutions to the Exercise, with graphs and step-by-step numericals. Maharashtra State Board (Balbharati), Std 10 Science and Technology Part 1.
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Exercise
(a) The amount of water vapour in air is determined in terms of its ……
The amount of water vapour in air is determined in terms of its absolute humidity.
(b) If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their ……
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacity.
(c) During transformation of liquid phase to solid phase, the latent heat is ……
During transformation of liquid phase to solid phase, the latent heat is given out (released).
Volume of 1 kg of water from 0 °C to 10 °C. The volume falls till 4 °C and then rises.
- Generally, a liquid expands when heated and contracts when cooled.
- But the graph shows that when water is heated from 0 °C to 4 °C, its volume decreases. Water contracts instead of expanding.
- At 4 °C the volume of water is minimum, so its density is maximum.
- Above 4 °C, water expands like other liquids and its volume increases.
This special behaviour of water between 0 °C and 4 °C is called the anomalous behaviour of water.
Specific heat capacity: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object. It is denoted by ‘c’. SI unit: J/kg °C; CGS unit: cal/g °C.
Experiment:
- Material: a tray with a thick layer of wax, solid spheres of iron, copper and lead of equal mass, a burner and a large beaker.
- Put all three spheres in boiling water for some time, so that all of them reach 100 °C.
- Take them out and put them immediately on the thick slab of wax.
- Note the depth to which each sphere sinks into the wax.
Observation: The iron sphere goes deepest into the wax, the copper sphere goes to intermediate depth and the lead sphere goes the least.
Conclusion: All three spheres had equal mass and the same rise in temperature, yet they absorbed different amounts of heat. The sphere that absorbed more heat gives more heat to the wax, melts more wax and goes deeper. So iron has the highest specific heat, then copper, and lead the least. This proves that different substances have different specific heat capacities.
The temperature interval chosen is 14.5 °C to 15.5 °C.
One calorie is the heat needed to raise the temperature of 1 g of water by 1 °C, from 14.5 °C to 15.5 °C.
Reason: If water is heated by 1 °C in a different temperature range, the heat required is slightly different. So a fixed temperature range must be stated to define the unit of heat exactly. (1 cal = 4.18 J)
AB: ice melts at 0 °C | BC: water warms from 0 °C to 100 °C | CD: water boils at 100 °C.
The graph shows what happens when ice is heated continuously.
- Line AB (0 °C): Ice melts into water. Heat is absorbed, but the temperature stays constant at 0 °C (melting point) until all the ice melts. This heat is the latent heat of fusion. Here ice and water are both present.
- Line BC (0 °C to 100 °C): After all the ice has melted, the heat raises the temperature of the water from 0 °C to 100 °C.
- Line CD (100 °C): Water boils. Heat is absorbed, but the temperature stays constant at 100 °C (boiling point) until all the water turns into vapour. This heat is the latent heat of vaporization. Here boiling water and vapour are both present.
So during a change of state, heat is absorbed but the temperature does not rise. The heat is used to weaken the bonds between the molecules.
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(a) What is the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
- In cold regions, the water at the top of a lake cools first. Down to 4 °C it becomes denser and sinks.
- When the top water cools below 4 °C, it expands, becomes lighter and stays at the top. It freezes into a layer of ice at 0 °C.
- Ice is a poor conductor of heat, so it stops the water below from losing heat.
- The water at the bottom remains at about 4 °C. So aquatic plants and animals survive even when the air temperature is below 0 °C.
(b) How can you relate the formation of water droplets on the outer surface of a bottle taken out of refrigerator with formation of dew?
- The cold bottle cools the air in contact with its outer surface.
- When this air cools below its dew point temperature, it becomes saturated with water vapour.
- The excess vapour condenses into tiny water droplets on the bottle.
- Dew forms in the same way. In the early morning, leaves, grass and window glass are cold. The air near them cools below the dew point and the vapour condenses into droplets called dew.
(c) In cold regions in winter, the rocks crack due to anomalous expansion of water.
- Water collects in the cracks of rocks.
- In winter, when this water cools below 4 °C, it expands instead of contracting. On freezing, its volume increases further.
- The expanding water and ice exert a very large pressure on the walls of the cracks.
- Due to this pressure, the rocks crack. For the same reason, water supply pipes burst in cold regions.
(a) What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Latent heat: The heat energy absorbed or given out at constant temperature during a change of state of a substance is called latent heat. It does not raise the temperature, so it is ‘hidden’ heat.
If latent heat is given off:
- a gas changes into a liquid (condensation), e.g. steam into water at 100 °C;
- a liquid changes into a solid (freezing), e.g. water into ice at 0 °C.
(b) Which principle is used to measure the specific heat capacity of a substance?
The principle of heat exchange is used: heat energy lost by the hot object = heat energy gained by the cold object. The specific heat is measured by the mixing method, using a calorimeter.
(c) Explain the role of latent heat in the change of state of a substance.
- During a change of state, the heat absorbed does not raise the temperature. It is used to weaken or break the bonds between the atoms or molecules.
- For example, the latent heat of fusion converts solid ice into liquid water at 0 °C. The latent heat of vaporization converts water into vapour at 100 °C.
- In the reverse changes (vapour to water, water to ice), the same latent heat is given out.
- So latent heat is what makes the change of state possible at a constant temperature.
(d) On what basis and how will you determine whether air is saturated with vapour or not?
Basis: Relative humidity.
- If the relative humidity is 100%, the air is saturated with vapour. The air is then at its dew point temperature.
- If the relative humidity is less than 100%, the air is unsaturated.
Practically, if water droplets form on a surface as soon as it is slightly cooler than the air, the air is at or very near its dew point, i.e. it is saturated.
(i) Heat is transferred from where to where?
Heat is transferred from the hot object to the cold object, until both attain the same temperature.
(ii) Which principle do we learn about from this process?
We learn the principle of heat exchange.
(iii) How will you state the principle briefly?
If a system of a hot and a cold object is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(iv) Which property of the substance is measured using this principle?
The specific heat capacity of a substance is measured using this principle.
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(a) Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3 °C and B by 5 °C. Which object has more specific heat? And by what factor?
Given: mA = mB = 1 g, ΔTA = 3 °C, ΔTB = 5 °C, QA = QB = Q.
For A: Q = 1 × cA × 3 For B: Q = 1 × cB × 5
∴ 3cA = 5cB ⇒ cAcB = 53
Object A has more specific heat, by a factor of 5/3. (A smaller rise in temperature for the same heat means a larger specific heat.)
(b) Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg ice at 0 °C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia = 341 cal/g)
Given: m = 2 kg = 2000 g of water, c = 1 cal/g °C, ΔT = 20 − 0 = 20 °C, latent heat of fusion of ice = 80 cal/g, latent heat of vaporization of ammonia = 341 cal/g.
Heat removed to freeze water at 0 °C into ice = m L = 2000 × 80 = 160000 cal
Total heat removed = 40000 + 160000 = 200000 cal
This heat is absorbed by evaporating ammonia: mNH3 × 341 = 200000
mNH3 = 200000341 = 586.5 g
About 586.5 g of ammonia must be evaporated.
The textbook answer is printed as 586.4 g. The exact value of 200000 ÷ 341 is 586.51, so 586.5 g is correct; both are accepted.
(c) A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to be mixed to it, so that water of temperature 50 °C will be obtained? (Given: latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Given: mice = 150 g at 0 °C, steam at 100 °C, final temperature = 50 °C. Let the mass of steam be m g.
to melt at 0 °C = 150 × 80 = 12000 cal
to heat water from 0 °C to 50 °C = 150 × 1 × 50 = 7500 cal
Total heat gained = 19500 cal
Heat lost by steam:
to condense at 100 °C = m × 540 ; to cool water from 100 °C to 50 °C = m × 1 × 50
Total heat lost = 540m + 50m = 590m
By the principle of heat exchange, 590m = 19500
m = 19500590 = 33.05 ≈ 33 g
About 33 g of steam has to be mixed.
(d) A calorimeter has mass 100 g and specific heat 0.1 kcal/kg °C. It contains 250 gm of liquid at 30 °C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0 °C, what will be the temperature of the mixture?
Given: calorimeter: 100 g, c = 0.1 kcal/kg °C = 0.1 cal/g °C; liquid: 250 g, c = 0.4 cal/g °C, at 30 °C (calorimeter also at 30 °C); ice: 10 g at 0 °C, latent heat of fusion = 80 cal/g. Let the final temperature be T °C.
Heat lost by liquid = 250 × 0.4 × (30 − T) = 100(30 − T)
Heat lost by calorimeter = 100 × 0.1 × (30 − T) = 10(30 − T)
Heat gained = Heat lost: 800 + 10T = 110(30 − T) = 3300 − 110T
120T = 2500 ⇒ T = 2500120 = 20.8 °C
The temperature of the mixture is 20.8 °C.
1 kcal/kg °C = 1 cal/g °C, so the specific heats can be used directly with masses in grams.
Making the model:
- Take a tall transparent plastic bottle (cut off its top) as the cylindrical container.
- Fix a wide plastic bowl or cup around the middle of the bottle as the outer bowl.
- Make two small holes in the bottle, one above the bowl and one below it. Fix thermometers T2 (upper) and T1 (lower) in them and seal with clay or glue.
- Fill the bottle with water at room temperature. Fill the bowl with a freezing mixture of crushed ice and salt.
Experiment: Record T1 and T2 every 30 seconds and plot both against time on the same graph.
Expected result:
- T1 (lower) falls quickly to 4 °C and stays there for some time, because water at 4 °C is densest and sinks.
- T2 (upper) falls slowly to 4 °C, then drops quickly to 0 °C, because water colder than 4 °C is lighter and rises.
- The two curves meet at 4 °C, the temperature of maximum density of water.