Matter in Our Surroundings — Practice

Question 1

Define 'matter' in one sentence and give two examples that are NOT matter.

Accepted Answer

Step 1 — Definition.
  Matter is anything that has mass and occupies space (volume).

Step 2 — Examples of NON-matter.
  • Light (a form of energy, no rest mass).
  • Sound (a wave, an energy disturbance, not a substance).
  • Heat (energy in transit between bodies).

✦ Answer: Matter has mass and occupies space; light and sound are NOT matter — they are forms of energy.

Question 2

Convert 27 °C to Kelvin.

Accepted Answer

Step 1 — Use T(K) = T(°C) + 273.15.

Step 2 — Substitute.
  T = 27 + 273.15 = 300.15 K.

✦ Answer: 300.15 K (often written 300 K to 3 s.f.).

Memo tip: room temperature in Kelvin is about 300 K.

Question 3

Convert 373 K to Celsius.

Accepted Answer

Step 1 — Use T(°C) = T(K) − 273.15.

Step 2 — Substitute.
  T = 373 − 273.15 = 99.85 °C ≈ 100 °C.

✦ Answer: ≈ 100 °C.

Context: 373 K is the boiling point of water — that's why this number appears so often.

Question 4

Why does the smell of hot food reach you faster than that of cold food?

Accepted Answer

Step 1 — Recall: at higher temperature, particles have higher kinetic energy and move faster.

Step 2 — Apply to diffusion.
  Hot food → faster-moving aroma particles → quicker diffusion through the air → smell reaches you sooner.
  Cold food → slower particles → slower diffusion.

✦ Answer: Higher temperature → faster particle motion → faster diffusion of aroma particles → smell reaches you more quickly.

This is a classic 2-mark CBSE question. Always tie temperature to kinetic energy in your answer.

Question 5

Name the process when (a) iodine crystals are heated and turn directly to vapour, (b) water vapour in air turns into dew on a cold surface.

Accepted Answer

Step 1 — Identify the from-state and to-state in each.

Step 2 — Match to the standard names.
  (a) Solid → Gas (no liquid) = SUBLIMATION.
  (b) Gas → Liquid = CONDENSATION.

✦ Answer: (a) Sublimation, (b) Condensation.

Direction-changing terms: gas-to-solid is DEPOSITION (e.g., frost on a winter window), and liquid-to-solid is FREEZING / SOLIDIFICATION.

Question 6

List two differences between liquids and gases at the particle level.

Accepted Answer

Step 1 — Compare on inter-particle distance.
  Liquid: small (just slightly bigger than solid). Gas: very large.

Step 2 — Compare on force of attraction.
  Liquid: moderate, enough to hold a definite volume. Gas: very weak, particles fly free.

(Optional third) Compressibility — Gas highly compressible, liquid almost incompressible — proves the space difference.

✦ Answer: (a) Particles in a gas are much further apart than in a liquid; (b) attractive forces between liquid particles are far stronger than in a gas (so liquids have a fixed volume; gases don't).

Question 7

How much heat is required to convert 200 g of ice at 0 °C completely to water at 0 °C? (L_f of ice = 334 kJ/kg)

Accepted Answer

Step 1 — Identify the process.
  Ice at 0 °C → Water at 0 °C is melting at the melting point. Temperature doesn't change; use Q = mL.

Step 2 — Convert mass to kg.
  m = 200 g = 0.2 kg.

Step 3 — Substitute.
  Q = m × L_f = 0.2 × 334 kJ = 66.8 kJ.

Step 4 — Convert to joules if needed.
  Q = 66.8 × 1000 = 66,800 J = 6.68 × 10⁴ J.

✦ Answer: Q = 66.8 kJ = 66,800 J.

Key idea: during a phase change at the phase-change temperature, all input heat goes into breaking bonds — none of it raises temperature.

Question 8

Calculate the heat released when 50 g of steam at 100 °C condenses to water at 100 °C. (L_v of water = 2260 kJ/kg)

Accepted Answer

Step 1 — Identify the process.
  Steam (gas) at 100 °C → Water (liquid) at 100 °C is condensation. Use Q = mL_v.
  Energy is RELEASED to the surroundings.

Step 2 — Convert mass.
  m = 50 g = 0.05 kg.

Step 3 — Substitute.
  Q = m × L_v = 0.05 × 2260 kJ = 113 kJ.

✦ Answer: 113 kJ of heat is released.

This is why steam burns are so severe: 1 kg of steam condensing on your skin releases 2260 kJ — about the energy of ten chocolate bars — entirely as heat into your skin.

Question 9

Explain at the particle level why a wet earthen pot keeps water inside cooler than a metal pot does.

Accepted Answer

Step 1 — Recall the cooling mechanism.
  Evaporation removes the highest-energy particles from a liquid surface, lowering the average kinetic energy (i.e. temperature) of what remains.

Step 2 — Apply to the pot.
  Earthen pots have tiny pores; water slowly seeps out and forms a thin film on the outer surface.
  This film evaporates, drawing heat away from the pot's outer wall.
  The wall in turn pulls heat from the water inside.
  Net effect: water inside cools below the surrounding air temperature.

Step 3 — Why a metal pot fails.
  Metal is non-porous; water cannot seep through. No outer-surface evaporation → no cooling.

✦ Answer: Earthen pot is porous → water seeps through → evaporates from the outer surface → removes heat from the pot → water inside cools. Metal pot is non-porous, so this mechanism doesn't operate.

Question 10

Why does the temperature of water remain at 100 °C while it is being boiled even though heat is continuously supplied?

Accepted Answer

Step 1 — At boiling point, the system is in a phase-transition.
  Liquid (water) is converting to gas (steam).

Step 2 — Where does the heat go?
  The heat is used as LATENT HEAT of vaporisation to break the inter-molecular bonds between water particles, NOT to raise temperature.

Step 3 — When does the temperature rise again?
  Only after all the water has converted to steam — once the phase change is complete, additional heat raises the temperature of the steam.

✦ Answer: During boiling, supplied heat is consumed as latent heat to break inter-particle bonds (liquid → gas). Until all water has vaporised, the temperature stays fixed at the boiling point (100 °C at 1 atm).

Question 11

How much heat is needed to convert 100 g of ice at 0 °C to water at 50 °C? (L_f = 334 kJ/kg, c_water = 4.2 kJ/kg·°C)

Accepted Answer

Step 1 — Break into two sub-processes.
  (i) Melt the ice (0 °C ice → 0 °C water): Q₁ = m × L_f.
  (ii) Heat the water (0 °C → 50 °C): Q₂ = m × c × ΔT.

Step 2 — Convert mass.
  m = 100 g = 0.1 kg.

Step 3 — Compute Q₁.
  Q₁ = 0.1 × 334 = 33.4 kJ.

Step 4 — Compute Q₂.
  Q₂ = 0.1 × 4.2 × (50 − 0) = 0.1 × 4.2 × 50 = 21 kJ.

Step 5 — Total.
  Q = Q₁ + Q₂ = 33.4 + 21 = 54.4 kJ.

✦ Answer: 54.4 kJ.

Lesson: split mixed problems into pure-phase-change and pure-temperature-change segments.

Question 12

Wet clothes dry faster on a windy summer afternoon than on a calm winter morning. Give two distinct reasons.

Accepted Answer

Step 1 — List the four factors affecting evaporation: surface area, temperature, humidity, wind speed.

Step 2 — Compare windy summer vs calm winter for each factor.
  (a) Temperature: summer afternoon is HIGHER. Higher T → more particles have escape energy → faster evaporation.
  (b) Wind speed: windy day. Wind carries away vapour from above the wet cloth → maintains a steep concentration gradient → faster evaporation.
  (c) Humidity is also typically lower on hot windy days (vapour gets carried away) → also helps. Mentioning any 2 is enough.

✦ Answer: (i) Higher temperature → more high-energy particles can escape; (ii) faster wind → vapour is swept away, allowing more to escape. Both increase evaporation rate.

Question 13

Calculate the total heat required to convert 50 g of ice at −10 °C to steam at 100 °C. Given: c_ice = 2.1 kJ/kg·°C, L_f = 334 kJ/kg, c_water = 4.2 kJ/kg·°C, L_v = 2260 kJ/kg.

Accepted Answer

Step 1 — Identify the four sub-processes.
  (i) Heat ice from −10 °C to 0 °C: Q₁ = m c_ice ΔT₁.
  (ii) Melt ice at 0 °C: Q₂ = m L_f.
  (iii) Heat water from 0 °C to 100 °C: Q₃ = m c_water ΔT₃.
  (iv) Vaporise water at 100 °C: Q₄ = m L_v.

Step 2 — Convert mass.
  m = 50 g = 0.05 kg.

Step 3 — Calculate each piece.
  Q₁ = 0.05 × 2.1 × 10 = 1.05 kJ.
  Q₂ = 0.05 × 334 = 16.7 kJ.
  Q₃ = 0.05 × 4.2 × 100 = 21.0 kJ.
  Q₄ = 0.05 × 2260 = 113.0 kJ.

Step 4 — Sum.
  Q_total = 1.05 + 16.7 + 21.0 + 113.0 = 151.75 kJ.

✦ Answer: 151.75 kJ ≈ 152 kJ.

Observation: Q₄ (vaporisation) dwarfs the others — vaporisation latent heat is by far the largest energy step in turning ice to steam.

Question 14

Explain why CNG (compressed natural gas) is sold compressed under high pressure but petrol is sold at atmospheric pressure.

Accepted Answer

Step 1 — Recall compressibility differences.
  Gases (like methane, the main component of CNG) have huge inter-particle space and are highly compressible.
  Liquids (like petrol) have tiny inter-particle space and are essentially incompressible.

Step 2 — Apply to storage.
  • CNG: at atmospheric pressure, the energy density is too low to be useful as fuel (a tank full of gas at 1 atm has very little mass of fuel). Compressing it to 200 atm packs ~200× more fuel into the same tank. This is feasible because gas particles can be pushed close together.
  • Petrol: it's already a dense liquid at atmospheric pressure — particles are essentially touching. Compressing further gives almost no extra mass per litre; the costs (stronger tank, special valves) outweigh negligible benefit.

Step 3 — Practical takeaway.
  Compressing a gas dramatically increases its useful mass per unit volume (energy density per litre). Compressing a liquid barely changes anything.

✦ Answer: Gases are highly compressible (large inter-particle spaces), so high-pressure storage in cylinders gives CNG a usable energy density. Liquids are nearly incompressible — compressing petrol gives no meaningful benefit.

Question 15

When dry ice (solid CO₂) is exposed to room conditions it forms a 'mist' a few centimetres above its surface — yet dry ice itself doesn't melt to a liquid. Explain both observations.

Accepted Answer

Step 1 — Why dry ice doesn't melt.
  At atmospheric pressure, solid CO₂ sublimes directly to gas (no liquid phase). Liquid CO₂ exists only above 5.1 atm of pressure (its triple point). Under normal conditions, the path solid→liquid is closed.

Step 2 — Why a 'mist' appears.
  As CO₂ vapour leaves the dry ice it is very cold (≈ −78.5 °C). This cold gas chills the air immediately above the dry ice. The air contains water vapour; chilling it below its dew point causes that WATER vapour to condense into tiny liquid water droplets — visible as mist/fog.

Step 3 — Note: the 'mist' is NOT CO₂ — it's WATER from the surrounding air.
  CO₂ gas itself is colourless and invisible. The mist is condensed water droplets caused by the cooling effect of CO₂ vapour.

✦ Answer: Dry ice sublimes (solid → gas) at atmospheric pressure because liquid CO₂ doesn't exist below ~5 atm. The cold CO₂ vapour chills the surrounding humid air, causing water vapour in the AIR (not CO₂) to condense into a visible mist of tiny water droplets.

Question 16

At higher altitude (e.g. on a mountain), why does water boil at a temperature lower than 100 °C?

Accepted Answer

Step 1 — Recall the definition of boiling point.
  Liquid boils when its vapour pressure equals the surrounding (atmospheric) pressure.

Step 2 — How altitude changes atmospheric pressure.
  Atmospheric pressure DECREASES with altitude (less air above pushing down). At sea level: 1 atm. On Mount Everest: ≈ 0.33 atm.

Step 3 — Conclude.
  Lower outside pressure → liquid needs less vapour pressure to boil → boiling happens at a lower temperature.
  E.g., water boils at ~71 °C on Everest, ~93 °C in Shimla, but 100 °C at the coast.

Step 4 — Practical consequence.
  At lower boiling points, food takes longer to cook (less heat per unit time goes into the food). This is why pressure cookers exist — they raise the boiling point above 100 °C by raising the pressure inside the cooker.

Practice — Matter in Our Surroundings

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities