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Physics Class 10 Chapter 10: Heat Capacity and Modes of Heat Transfer

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This is the free study page for Physics Class 10, Chapter 10: Heat Capacity and Modes of Heat Transfer. It follows the Federal Board (FBISE) SSC-II syllabus. This chapter has 17 learning outcomes, and Parhayi.com has 205 MCQs, 52 short questions and 15 long questions for it. Read the topics and the key questions with their answers below, then tap Practice this chapter to try the questions yourself.

Physics explains how the world works, from motion and forces to heat, light and magnets. Learn each idea in simple steps, check the units, and practise numericals every day.

What you will learn

Key questions and answers

These are the short questions of this chapter with their answers (52 questions).

Q1. a. How do increased temperatures lead to more frequent heatwaves in cities? (3 marks)

A heat wave is a period of abnormally hot weather lasting more than two days. Global warming raises the average temperature, so the hot days become hotter and more frequent. Cities are affected most because roads, concrete and dark roofs absorb more solar radiation than grass or water and emit it slowly at night (urban heat island); there are fewer trees for shade and evaporation, and vehicles, air-conditioners and factories add heat and greenhouse gases. So the extra warming pushes city temperatures above the heat-wave level more often.

Q2. b. How might the intensity of future hurricanes change if global temperatures continue to rise? (3 marks)

Hurricanes get their energy from warm ocean water. Warmer oceans mean more evaporation and more latent heat released when the vapour condenses, so the storms become stronger, with faster winds and heavier rain. Warm air also holds more moisture, so hurricanes will drop even more rain and cause more flooding. Future hurricanes are therefore expected to be more intense.

Q3. c. Explain what the glass roof of a greenhouse and carbon dioxide in Earth's atmosphere have in common in terms of heat and temperature. (3 marks)

Both let the short-wave radiation of the Sun pass in, but do not let the long-wave infrared radiation emitted by the warm surface inside escape easily. The glass roof traps the infrared radiation inside the greenhouse so plants stay warm; carbon dioxide (and other greenhouse gases) absorbs the infrared radiation emitted by the Earth and re-emits it back, keeping the Earth's surface warmer than it would otherwise be. In both cases heat is trapped and the temperature rises.

Q4. b. Which heats up quickly: water or metal? (3 marks)

Iron (the metal). In 60 minutes iron rises by 40 °C (25 → 65 °C) while water rises by only 5 °C (25 → 30 °C). The iron line is much steeper. Reason: specific heat of iron (452 J kg⁻¹ K⁻¹) is about nine times smaller than that of water (4200 J kg⁻¹ K⁻¹), so the same heat gives it a bigger temperature rise.

Q5. c. Which will cool more slowly: water or iron? (3 marks)

Water. Because of its large specific heat, water stores a large amount of heat for every degree of temperature; to cool by 1 °C it must lose 4200 J per kg, while iron loses only 452 J per kg. So water takes much longer to cool – a hot cup of tea stays hot longer than a hot spoon.

Q6. d. When you boil water in an iron pot on the stove, which heats up faster: the iron pot or the water? (3 marks)

The iron pot heats faster. It is in direct contact with the flame, it is a good conductor, and its specific heat is small, so its temperature rises quickly; the water needs about nine times more heat per kilogram for the same rise, and it receives the heat only from the pot. (This is why an empty pot on the stove becomes red hot very quickly.)

Q7. Why should we wear dark-coloured clothes in winter and white-coloured clothes in summer? (3 marks)

Dark (black) surfaces are good absorbers of radiation. In winter dark clothes absorb the Sun's radiation and keep us warm. White surfaces are bad absorbers and good reflectors, so in summer white clothes reflect most of the sunlight and keep the body cool.

Q8. In a house, geysers are fitted on the ground floor, and still we get warm water on the top floor without a pump. How is it possible? (3 marks)

By convection. The water heated in the geyser expands, becomes less dense and rises through the pipes to the upper floor, while the colder, denser water flows down to the geyser to be heated. This natural convection current keeps hot water flowing upward without any pump.

Q9. Where will you get more heat from a wood fire, 1 metre above the wood or 1 metre in front of it? (3 marks)

One metre above the fire. In front of the fire we receive heat only by radiation. Above the fire we receive the radiation plus the hot air and gases rising by convection, so the heat received is more.

Q10. Why do crowded city areas feel hotter compared to the outskirts on a hot summer day? (3 marks)

(i) Roads, concrete buildings and dark roofs absorb more solar radiation than soil, grass and trees and have low specific heat, so they heat up quickly. (ii) There are few trees and water bodies, so there is less cooling by shade and evaporation. (iii) Tall buildings block the wind and stop convection currents. (iv) Vehicles, air-conditioners and factories release extra heat and greenhouse gases. All this makes the city an "urban heat island".

Q11. Why is the metallic handle of a door colder than the wood of the same door when touched? (3 marks)

Both are at the same room temperature, but metal is a good conductor of heat (free electrons) while wood is a bad conductor. When touched, the metal conducts heat away from the hand very quickly, so the hand loses heat fast and the metal feels cold. Wood conducts heat very slowly, so it does not feel cold.

Q12. How do trees help reduce the effects of climate change, and what could happen if forests are depleted? (3 marks)

Trees absorb carbon dioxide (a greenhouse gas) from the air during photosynthesis and store carbon; they also give shade, release water vapour that cools the air, and hold the soil. If forests are cut down, less CO₂ is absorbed and the carbon stored in the trees is released, so greenhouse gases increase, global warming speeds up, and floods, droughts, heat waves and soil erosion become worse.

Q13. Why do certain gases in the atmosphere trap more heat than others? (3 marks)

Gases such as water vapour, carbon dioxide and methane have molecules that can absorb infrared (heat) radiation emitted by the Earth and re-emit it in all directions, including back towards the Earth. Gases like nitrogen and oxygen do not absorb infrared radiation, so they let it escape. The absorbing gases are the greenhouse gases and they trap heat.

Q14. How does the specific heat capacity of different materials affect their use in cookware? (3 marks)

Metals with low specific heat (copper, aluminium, steel) heat up quickly with little heat, so pans and tawas are made from them and cooking starts fast. Materials with high specific heat or low conductivity (water, clay, wood, plastic) heat slowly and keep heat longer – so clay handis keep food warm, water is used for even cooking, and handles are made of plastic or wood so they stay cool.

Q15. Convert the specific heat of water, 4180 J kg⁻¹ K⁻¹, into J g⁻¹ °C⁻¹. (3 marks)

StepReason
Given: c = 4180 J kg−1 K−1Write the value we must convert.
1 kg = 1000 g and 1 K change = 1 °C changeThese two facts give the conversion of the mass and temperature units.
c = 4180 J1000 g × 1 °CReplace kg by 1000 g and K by °C.
c = 41801000 = 4.18Divide the number by 1000 to get the new value.
c = 4.18 J g−1 °C−1

Q16. Calculate the heat given to 25 kg of water to increase its temperature by 50 °C. (3 marks)

StepReason
m = 25 kg, c = 4200 J kg−1 K−1, ΔT = 50 °CWrite the given data.
Q = m c ΔTHeat gained depends on mass, specific heat and temperature change.
Q = 25 × 4200 × 50Put the values in the formula.
Q = 5 250 000 J = 5.25 × 106 JMultiply to get the heat; 1 MJ = 106 J.
Q = 5.25 × 106 J = 5.25 MJ

Q17. A half kg block of an unknown metal is heated from 30 °C to 80 °C, requiring 19 kJ of heat. (a) Find the specific heat of the metal. (b) How much additional heat is needed to heat it from 80 °C to 125 °C? (3 marks)

StepReason
m = 0.5 kg, Ti = 30 °C, Tf = 80 °C, Q = 19 000 JWrite the given data for part (a).
ΔT = Tf − Ti = 80 − 30 = 50 °CFind the temperature change.
c = Qm ΔTQ = m c ΔT is rearranged to find c.
c = 19 0000.5 × 50 = 19 00025 = 760 J kg−1 K−1Put the values in and divide; this is the answer of (a).
(b) ΔT = 125 − 80 = 45 °CFor part (b) the new temperature change is needed.
Q = m c ΔTUse the same heat formula again, now with c known.
Q = 0.5 × 760 × 45 = 17 100 JPut the values in and multiply.
(a) c = 760 J kg−1 K−1; (b) Q = 17 100 J = 17.1 kJ

Q18. Calculate the change in temperature of 5 litres of water if it absorbs 8.4 MJ of heat (c = 4200 J kg⁻¹ K⁻¹). (3 marks)

StepReason
5 litres of water: m = 5 kg, c = 4200 J kg−1 K−1, Q = 8.4 MJ = 8.4 × 106 J1 litre of water has a mass of 1 kg, and MJ is changed to J.
Q = m c ΔT, so ΔT = Qm cRearrange the heat formula to get the temperature change.
ΔT = 8.4 × 1065 × 4200 = 8.4 × 10621 000Put the values in the formula.
ΔT = 400 °CDivide to get the temperature change.
ΔT = 400 °C

Q19. 100 g of water at 80 °C is mixed with 200 g of water at 20 °C. Find the final temperature (no heat loss). (3 marks)

StepReason
Hot water: m1 = 100 g, T1 = 80 °C; cold water: m2 = 200 g, T2 = 20 °C; final temperature = TWrite the given data; T is the unknown.
Heat lost by hot water = heat gained by cold waterWith no heat loss, all the heat given out is taken in.
m1 c (T1 − T) = m2 c (T − T2)Use Q = m c ΔT on both sides; c is the same for water and cancels.
100 (80 − T) = 200 (T − 20)Put the values in.
8000 − 100 T = 200 T − 4000Open the brackets.
12 000 = 300 TCollect the terms of T on one side.
T = 12 000300 = 40 °CDivide to find the final temperature.
Final temperature = 40 °C

Q20. A 200 g solid at 110 °C is placed in a calorimeter (mass 60 g, c = 0.9 J g⁻¹ °C⁻¹) containing 200 g water at 20 °C. The final temperature is 35 °C. Find the specific heat of the solid. (3 marks)

StepReason
Solid: ms = 200 g, T = 110 °C; water: mw = 200 g, cw = 4.18 J g−1 °C−1, 20 °C; calorimeter: mc = 60 g, cc = 0.9 J g−1 °C−1; final Tf = 35 °CWrite the given data.
Heat lost by solid = heat gained by water + heat gained by calorimeterWith no heat loss to the surroundings, heat lost equals heat gained.
Q = m c ΔTThis formula gives the heat for each body.
Qwater = 200 × 4.18 × (35 − 20) = 12 540 JHeat gained by the water.
Qcal = 60 × 0.9 × (35 − 20) = 810 JHeat gained by the calorimeter, which also warms up.
Qgained = 12 540 + 810 = 13 350 JAdd the two heats gained.
Qlost = 200 × cs × (110 − 35) = 15 000 csHeat lost by the solid with unknown cs.
15 000 cs = 13 350Equate heat lost and heat gained.
cs = 13 35015 000 = 0.89Divide to find the specific heat of the solid.
cs = 0.89 J g−1 °C−1 (= 890 J kg−1 K−1)

Q21. A hot steel rod (mass 3 kg, initially 450 °C) is plunged into cold water and cools to 50 °C. Specific heat of steel = 460 J kg⁻¹ °C⁻¹. Find the heat lost by the rod. (3 marks)

StepReason
m = 3 kg, c = 460 J kg−1 °C−1, Ti = 450 °C, Tf = 50 °CWrite the given data.
ΔT = 450 − 50 = 400 °CFind the fall in temperature.
Q = m c ΔTHeat lost depends on mass, specific heat and temperature change.
Q = 3 × 460 × 400Put the values in the formula.
Q = 552 000 JMultiply to get the heat lost.
Q = 5.52 × 105 J = 552 kJ

Q22. Why do coastal areas have moderate temperatures than inland areas? Explain briefly. (3 marks)

Water has a very large specific heat capacity (4200 J kg⁻¹ K⁻¹) compared with soil and rock (about 800 J kg⁻¹ K⁻¹). The sea therefore warms up slowly in the day and cools slowly at night, so it keeps the nearby land from becoming too hot or too cold.

Also, during the day cool air moves from the sea to the land (sea breeze), and at night air moves from the land to the sea (land breeze). Inland areas have no such large store of heat, so their temperature swings are large.

Q23. How do sea and land breezes help in moderating coastal climate? (3 marks)

Sea breeze (day): Land has a small specific heat and gets hot quickly. The hot air above it rises and cool, moist air from the sea moves in to replace it. This cools the coast.

Land breeze (night): Land cools faster than the sea. The air above the warm sea rises and cooler air from the land moves out to sea. This keeps the night temperature mild.

Q24. How do land and sea breezes form, and what role does the temperature difference between land and sea water play in these breezes? (3 marks)

Land heats and cools faster than water. In the day the land is hotter, so air above it rises and creates a low-pressure area; cooler air from the sea (high pressure) flows in: a sea breeze. At night the sea is warmer, air above it rises, and air flows from the land to the sea: a land breeze.

The temperature difference between land and sea causes the density and pressure difference that drives the wind.

Q25. How does convection in seawater support marine life? (3 marks)

Convection currents in the sea are caused by differences in temperature and salinity. Upwelling lifts cold, nutrient-rich water from the depths to the surface, where phytoplankton use it with sunlight and become food for fish. Downwelling carries oxygen-rich surface water to the deep sea so that deep-water animals can breathe. Convection also spreads heat around the oceans and keeps the temperature suitable for marine life.

Q26. Calculate the heat required to raise the temperature of a 2 kg substance from 40 °C to 45 °C, given specific heat = 420 J kg⁻¹ K⁻¹. (3 marks)

StepReason
m = 2 kg, c = 420 J kg−1 K−1, Ti = 40 °C, Tf = 45 °CWrite the given data.
ΔT = 45 − 40 = 5 °CFind the temperature rise.
Q = m c ΔTHeat required depends on mass, specific heat and temperature change.
Q = 2 × 420 × 5Put the values in the formula.
Q = 4200 JMultiply to get the heat.
Q = 4200 J

Q27. What is meant by specific heat capacity? Give its mathematical expression. (3 marks)

Specific heat capacity of a substance is the amount of heat required to change the temperature of 1 kg of the substance by 1 K (or 1 °C).

c = Q / (m ΔT) or Q = m c ΔT

Its SI unit is J kg⁻¹ K⁻¹.

Q28. What is meant by specific heat capacity of a substance? Calculate the amount of heat required to increase the temperature of 25 kg of water by 50 °C. (c = 4200 J kg⁻¹ K⁻¹) (3 marks)

StepReason
Specific heat capacity is the heat needed to raise the temperature of 1 kg of a substance by 1 K.This is the definition asked in the first part.
m = 25 kg, c = 4200 J kg−1 K−1, ΔT = 50 °CWrite the given data for the calculation.
Q = m c ΔTHeat needed depends on mass, specific heat and temperature change.
Q = 25 × 4200 × 50Put the values in the formula.
Q = 5 250 000 JMultiply to get the heat.
Q = 5.25 × 106 J

Q29. What is meant by specific heat capacity of a substance? Write its formula and SI unit. (3 marks)

Specific heat capacity is the heat absorbed or released by 1 kg of a substance for a temperature change of 1 K.

c = Q / (m ΔT)
SI unit: J kg⁻¹ K⁻¹

Q30. Define heat capacity of an object. How does it differ from specific heat capacity? (3 marks)

Heat capacity is the heat needed to raise the temperature of the whole object by 1 K: C = Q/ΔT = m c, unit J K⁻¹.

Specific heat capacity is for 1 kg of the material: c = Q/(mΔT), unit J kg⁻¹ K⁻¹. Heat capacity depends on the mass of the object; specific heat does not.

Q31. A 2 kg iron block at 150 °C is placed in 5 kg of water at 25 °C. Find the final temperature. (c for iron = 452 J kg⁻¹ K⁻¹, water = 4200 J kg⁻¹ K⁻¹; ignore heat losses.) (3 marks)

StepReason
Iron: m1 = 2 kg, c1 = 452, T1 = 150 °C; water: m2 = 5 kg, c2 = 4200, T2 = 25 °C (c in J kg−1 K−1); final temperature = TWrite the given data; T is the unknown.
Heat lost by iron = heat gained by waterHeat losses are ignored, so the two heats are equal.
m1 c1 (T1 − T) = m2 c2 (T − T2)Write Q = m c ΔT for each body.
2 × 452 × (150 − T) = 5 × 4200 × (T − 25)Put the values in.
904 (150 − T) = 21 000 (T − 25)Multiply the numbers on each side.
135 600 − 904 T = 21 000 T − 525 000Open the brackets.
660 600 = 21 904 TCollect the terms of T on one side.
T = 660 60021 904 = 30.2 °CDivide to find the final temperature.
Final temperature ≈ 30.2 °C

Q32. Why is a metal cooking pan fitted with a wooden or plastic handle? Name the heat transfer involved. (2 marks)

Heat from the flame passes through the metal pan to the handle by conduction. Wood and plastic are insulators, so the handle stays cool enough to hold, while the metal pan conducts heat quickly to the food.

Q33. Why are heat sinks in computers made of aluminium with many thin fins? (3 marks)

Aluminium is a good conductor, so heat flows quickly from the processor into the sink. The many fins give a large surface area, so more heat can be given to the air by convection and radiation. A fan can increase convection. Painting it black increases radiation.

Q34. Explain why water is used in a hot-water bottle and not oil. (2 marks)

Water has a large specific heat capacity (4200 J kg⁻¹ K⁻¹), more than oil. For the same mass and temperature it stores more heat energy and cools slowly, so the bottle keeps the bed warm for hours.

Q35. Describe the experiment to show that convection currents exist in water. (3 marks)

Fill a beaker with water and drop a few crystals of potassium permanganate at the bottom. Heat the beaker gently from below, just under the crystals. The purple colour rises in a stream above the flame, spreads across the top and sinks at the sides. This shows that the heated water, which is less dense, rises and the cooler, denser water sinks: a convection current.

Q36. Why does a wind blow from the land to the sea at night? (3 marks)

Land has a smaller specific heat than water, so it cools faster after sunset. The air above the sea is warmer, less dense and rises. This makes a low-pressure area over the sea, and cooler air from the land flows out to take its place: a land breeze.

Q37. Why can eagles glide for hours without flapping their wings? (2 marks)

The Sun heats dark ground more than grass or water, and the air above it becomes less dense and rises as a thermal. Eagles spread their wings and circle inside the rising air, which lifts them without effort. They glide to the next thermal when they need more height.

Q38. Why does a shiny surface emit less radiation than a dull black one at the same temperature? (2 marks)

A surface that reflects most of the radiation falling on it absorbs little. Emission and absorption go together, so it is also a poor emitter. A dull black surface absorbs most radiation and emits it equally well.

Q39. How does the rate of emission of radiation depend on temperature and surface area? (2 marks)

The hotter an object is, the more infrared radiation it emits every second. At the same temperature, a body with a larger surface area emits more radiation than one with a smaller surface area. This is why radiators have fins.

Q40. Describe the Leslie cube experiment and give its result. (3 marks)

A Leslie cube is a hollow metal cube with four faces: dull black, white, dark colour and shiny metal. It is filled with boiling water. An infrared thermometer is held at the same distance from each face. Result: the dull black face gives the highest reading (best emitter) and the shiny face gives the lowest (poorest emitter).

Q41. How does the greenhouse effect warm the Earth? Name two greenhouse gases. (3 marks)

Sunlight passes through the atmosphere and warms the ground. The ground emits infrared radiation. Greenhouse gases such as carbon dioxide and methane absorb part of this infrared and send it back, so the Earth stays warmer than it would otherwise be.

Q42. How does a thermos flask reduce heat transfer by all three modes? (3 marks)

A thermos has two glass walls with a vacuum between them. The vacuum has no particles, so there is no conduction or convection across the gap. The inner surfaces are silvered to reflect heat radiation. The stopper and plastic outer case are poor conductors and prevent heat loss through the top.

Q43. Why does the temperature of a place near the sea not change much during the day? (2 marks)

Water has a large specific heat, so it takes up a large amount of heat in the day for a small rise in temperature and gives it out slowly at night. Sea and land breezes also move heat.

Q44. How do cyclones form? Use ideas of convection. (3 marks)

Over warm ocean water, evaporation makes warm, moist air that rises by convection. As it rises it cools and the vapour condenses into clouds, releasing latent heat that warms the air and keeps it rising. The pressure near the sea surface falls, so more warm moist air rushes in. The Earth’s rotation makes the air spiral, forming a cyclone.

Q45. A 60 W lamp wastes about 50 W as heat. Explain how this heat would reach a person sitting near the lamp. (2 marks)

The lamp emits infrared radiation in all directions, which reaches the person directly (radiation). The air around the hot bulb rises by convection. Conduction occurs only through the lamp fittings and any air in contact.

Q46. Calculate the heat released when 5 kg of hot water cools from 90 °C to 30 °C. (c = 4200 J kg⁻¹ K⁻¹). Express in MJ. (3 marks)

StepReason
m = 5 kg, c = 4200 J kg−1 K−1, Ti = 90 °C, Tf = 30 °CWrite the given data.
ΔT = 90 − 30 = 60 °CFind the fall in temperature.
Q = m c ΔTHeat released depends on mass, specific heat and temperature change.
Q = 5 × 4200 × 60 = 1 260 000 JPut the values in and multiply.
Q = 1 260 000106 MJ = 1.26 MJChange joule into megajoule as asked.
Q = 1.26 MJ

Q47. A 0.3 kg aluminium pan (c = 900 J kg⁻¹ K⁻¹) containing 1 kg of water (c = 4200) is heated from 20 °C to 100 °C. Find the total heat absorbed. (3 marks)

StepReason
Pan: m1 = 0.3 kg, c1 = 900; water: m2 = 1 kg, c2 = 4200 (c in J kg−1 K−1); Ti = 20 °C, Tf = 100 °CWrite the given data.
ΔT = 100 − 20 = 80 °CBoth pan and water have the same temperature rise.
Q = m c ΔTThis formula is used for each body.
Qpan = 0.3 × 900 × 80 = 21 600 JHeat absorbed by the pan.
Qwater = 1 × 4200 × 80 = 336 000 JHeat absorbed by the water.
Q = 21 600 + 336 000 = 357 600 JAdd the two heats to get the total.
Total heat = 3.58 × 105 J

Q48. An electric heater of 2 kW is used to heat 10 kg of water (c = 4200) for 3 minutes. If 80% of the energy heats the water, find the temperature rise. (3 marks)

StepReason
P = 2 kW = 2000 W, t = 3 min = 180 s, m = 10 kg, c = 4200 J kg−1 K−1, efficiency = 80 %Write the given data in SI units.
E = P t = 2000 × 180 = 360 000 JEnergy supplied by the heater is power multiplied by time.
Q = 0.80 × 360 000 = 288 000 JOnly 80 % of this energy heats the water.
Q = m c ΔT, so ΔT = Qm cUse the heat formula and rearrange it for the temperature rise.
ΔT = 288 00010 × 4200 = 288 00042 000Put the values in the formula.
ΔT = 6.86 °CDivide to get the temperature rise.
ΔT ≈ 6.9 °C

Q49. A 0.8 kg copper kettle (c = 387 J kg⁻¹ K⁻¹) holds 1.5 kg of water at 25 °C. Calculate the total heat needed to bring both to 100 °C, and explain why the kettle's own heat is usually ignored. (3 marks)

StepReason
Kettle: m1 = 0.8 kg, c1 = 387 J kg−1 K−1; water: m2 = 1.5 kg, c2 = 4200 J kg−1 K−1; 25 °C → 100 °CWrite the given data; both are heated together.
ΔT = 100 − 25 = 75 °CBoth have the same temperature rise.
Q = m c ΔTThis formula is used for each body.
Qkettle = 0.8 × 387 × 75 = 23 220 JHeat taken by the kettle.
Qwater = 1.5 × 4200 × 75 = 472 500 JHeat taken by the water.
Q = 23 220 + 472 500 = 495 720 JAdd the two heats to get the total.
The kettle’s share is less than 5 %, because the specific heat of copper is about eleven times smaller than that of water.So for rough estimates the kettle’s own heat is ignored.
Total heat ≈ 4.96 × 105 J

Q50. A student says: 'A copper rod feels colder than a wooden rod in the same room, so the copper must be at a lower temperature.' Explain why this is wrong, using ideas of conduction. (3 marks)

Both rods have been in the room long enough to be at the same temperature. When touched, heat flows from the warmer hand into each rod. Copper has many free electrons and conducts this heat away very quickly, so the skin cools fast and the rod feels cold. Wood is a poor conductor, so the skin loses heat slowly and the rod feels warm. The sensation depends on the rate of heat flow, not on the actual temperature.

Q51. Explain, using density changes, why a room heater is placed near the floor but an air-conditioner is placed near the ceiling. (3 marks)

Air near a floor heater expands, becomes less dense and rises; cooler, denser air sinks to take its place and is heated in turn, so a convection current carries warmth through the whole room. An air-conditioner near the ceiling releases cold, dense air that sinks and pushes warmer air upward to be cooled, again setting up circulation. If the heater were high or the cooler low, the treated air would stay where it was and most of the room would be unaffected.

Q52. The Earth receives energy from the Sun yet its average temperature stays roughly constant. Explain this balance and how extra carbon dioxide disturbs it. (3 marks)

The Earth absorbs short-wave radiation from the Sun and, being much cooler, emits long-wave infrared radiation to space. When the rate of emission equals the rate of absorption, the average temperature stays constant. Greenhouse gases such as CO₂ let sunlight pass but absorb much of the outgoing infrared and re-radiate part of it back to the surface. Adding CO₂ reduces the net energy escaping, so the Earth warms until its higher temperature increases emission enough to restore the balance: this is global warming.

Long questions with answers

Q1. Explain the concept of specific heat capacity. Discuss applications of water based on its high specific heat capacity. (5 marks)

Specific heat capacity of a material is the amount of heat required to raise the temperature of 1 kg of the substance by 1 K (1 °C): c = Q/(mΔT), SI unit J kg⁻¹ K⁻¹. Different materials need different amounts of heat: metals have low c (copper 387 J kg⁻¹ K⁻¹) and heat up or cool quickly; water has a very high c (4200 J kg⁻¹ K⁻¹) and heats and cools slowly, storing a large amount of heat. The heat capacity of a body is C = mc (J K⁻¹).

Applications of water's high specific heat: (1) Moderate coastal climate – the sea absorbs a lot of heat in summer and releases it slowly in winter, so Karachi's temperature varies only from about 19 °C to 32 °C while inland areas go from 0 °C to 50 °C. (2) Stable lakes and oceans – slow temperature changes protect aquatic life. (3) Body temperature – our body is 60 % water, which helps keep the temperature near 37 °C. (4) Coolant – water in car radiators, power plants and industrial heat exchangers removes large amounts of heat and prevents overheating. (5) Cooking and hot drinks – water heats evenly without burning food, and tea stays hot for a long time. (6) Hot-water bottles and central heating carry heat for hours.

Q2. Analyse heat transfer through radiation, explaining why it is the fastest method. What factors influence the rate of heat transfer by radiation? (5 marks)

Radiation is the transfer of heat in the form of electromagnetic waves, mainly infrared radiation. It needs no material medium – heat from the Sun crosses the vacuum of space to reach the Earth, and we feel the warmth of a fire from a distance. Every object above absolute zero emits infrared radiation; hotter objects emit more. Fastest method: conduction and convection depend on the slow movement of particles, but radiation is an electromagnetic wave travelling at the speed of light, 3 × 10⁸ m/s, so heat reaches the receiver almost instantly (sunlight takes only about 8 minutes to cover 150 million km).

Factors affecting the rate of radiation: (1) Temperature of the surface – the hotter the object, the more radiation it emits per second (the Sun radiates far more than a stove). (2) Surface area – at the same temperature a larger surface emits more; radiators and heat sinks have fins. (3) Colour and texture (nature) of the surface – dull black and rough surfaces are the best absorbers and emitters; white and shiny silver surfaces are poor emitters but good reflectors (Leslie cube and two-cans experiments). A body whose emission equals absorption stays at constant temperature (thermal equilibrium).

Q3. Explain greenhouse effect. Is it good or bad? Discuss the consequences of greenhouse effect as well. (5 marks)

Greenhouse effect: Sunlight passes through the atmosphere and warms the Earth. The warm Earth emits long-wavelength infrared radiation. Greenhouse gases (carbon dioxide, methane, water vapour) absorb part of it and send it back to the ground. This trapping of heat is the greenhouse effect.

Good or bad? The natural greenhouse effect is good: without it the Earth would be about 33 °C colder and life would be difficult. But the extra greenhouse gases from burning fuels and cutting forests make the effect stronger. This enhanced greenhouse effect is bad: it causes global warming.

Consequences: more heat waves; melting glaciers and polar ice; rise in sea level and coastal flooding; stronger hurricanes and heavier rainfall; droughts and wildfires; loss of habitats and damage to crops.

Q4. Describe the causes and consequences of global warming. How does it lead to extreme weather events? (5 marks)

Causes: Burning coal, oil and gas, and cutting forests, increase carbon dioxide and methane. These gases trap more of the infrared radiation from the Earth (enhanced greenhouse effect), so the average temperature rises.

Extreme weather: Higher temperature causes heat waves; the warmer air holds more water vapour, which gives heavy rainfall and flooding; hot, dry weather causes wildfires; warmer seas supply more energy to hurricanes and cyclones; melting ice and expansion of sea water cause sea-level rise and coastal flooding.

Solutions: use renewable energy, plant trees, save energy.

Practice MCQs

Choose the answer in your mind first. The answer is shown below each question.

1. Why is water used in radiators of automobiles as coolant?

  1. It is easily available
  2. It is low cost or free
  3. It has large specific heat
  4. It has oxygen

Answer: c

Water absorbs a large amount of heat from the engine for a small rise in its own temperature.

2. On a summer afternoon beach sand is very hot but sea water is cool. This is because sand has:

  1. a lower density than water
  2. a larger specific heat than water
  3. a smaller specific heat than water
  4. more mass than water

Answer: c

For the same heat, sand (c ≈ 800) warms about five times more than water (c = 4200).

3. Heat from the Sun reaches the Earth by:

  1. conduction
  2. convection
  3. evaporation
  4. radiation

Answer: d

Space is empty.

4. Which process drives the formation of a cyclone?

  1. Conduction through the sea bed
  2. Radiation from the Moon
  3. Convection of warm moist air
  4. Evaporation of rocks

Answer: c

Rising warm moist air creates low pressure.

5. Which one does NOT make a thermal stronger?

  1. A cloudy, cool day
  2. Dry ground
  3. Dark ploughed land
  4. Strong sunshine

Answer: a

Thermals need strongly heated ground.

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