Heat Transfer in Nature – Numerical Problems with Stepwise Solutions
Class 7
Science
Chapter 7
Heat Transfer in Nature — 20 Numerical Problems with Stepwise Solutions
Topic-wise numerical practice strictly aligned with NCERT for CBSE Class 7. Step-by-step solutions show working and units.
Content Bank — Important formulas & concepts
- Heat required to change temperature: Q = m × c × ΔT, where Q = heat (J), m = mass (kg), c = specific heat capacity (J/kg°C), ΔT = change in temperature (°C).
- Latent heat (phase change): Q = m × L, where L is latent heat (J/kg) — e.g., latent heat of vaporisation or fusion.
- Specific heat (common values used here): water = 4200 J/kg°C (approx), aluminium ≈ 900 J/kg°C, iron ≈ 450 J/kg°C (use only when stated).
- Simple percentage/ratio for radiation absorption: Energy absorbed = incident energy × absorbance (fraction).
- Qualitative ideas: conduction (direct contact), convection (fluid motion), radiation (electromagnetic waves; can travel through vacuum).
1. (Conduction) How much heat is required to raise the temperature of 0.5 kg of water from 20°C to 60°C? (Take c = 4200 J/kg°C)
- Given: m = 0.5 kg, c = 4200 J/kg°C, ΔT = 60 − 20 = 40°C.
- Use Q = m × c × ΔT = 0.5 × 4200 × 40.
- Compute: 0.5 × 4200 = 2100; 2100 × 40 = 84,000 J.
- Answer: 84,000 J (or 84 kJ).
2. (Conduction) A 200 g piece of iron at 100°C is dropped into 500 g of water at 30°C. Assuming final temperature is 35°C and specific heat of iron = 450 J/kg°C, compute the heat lost by iron and gained by water. (Neglect heat loss to surroundings.)
- Convert masses: m_iron = 200 g = 0.200 kg; m_water = 500 g = 0.500 kg.
- Heat lost by iron: Q_iron = m × c × ΔT = 0.200 × 450 × (100 − 35) = 0.200 × 450 × 65.
- Compute: 0.200 × 450 = 90; 90 × 65 = 5,850 J (heat lost).
- Heat gained by water: Q_water = 0.500 × 4200 × (35 − 30) = 0.500 × 4200 × 5.
- 0.500 × 4200 = 2100; 2100 × 5 = 10,500 J (heat gained).
- Note: Since ideal heat balance is not strict with assumed final T, this shows computed values: iron gives 5,850 J, water gains 10,500 J — in real mixing the final T would lie where heat lost = heat gained.
3. (Conduction — Mixing) 250 g of water at 80°C is mixed with 250 g of water at 20°C. Find final temperature. (c for water = 4200 J/kg°C)
- m1 = 0.250 kg at T1 = 80°C; m2 = 0.250 kg at T2 = 20°C.
- Heat lost by hot water = heat gained by cold water → m1 c (T1 − Tf) = m2 c (Tf − T2).
- Since masses and c are equal, (T1 − Tf) = (Tf − T2) → 80 − Tf = Tf − 20.
- So 80 + 20 = 2 Tf → 100 = 2 Tf → Tf = 50°C.
- Answer: 50°C.
4. (Conduction) A 1.5 kg aluminium block is heated from 25°C to 75°C. (Take c_al = 900 J/kg°C.) Calculate heat absorbed.
- Given: m = 1.5 kg, c = 900 J/kg°C, ΔT = 50°C.
- Q = m c ΔT = 1.5 × 900 × 50.
- Compute: 1.5 × 900 = 1350; 1350 × 50 = 67,500 J.
- Answer: 67,500 J (67.5 kJ).
5. (Conduction) A kettle contains 1.0 kg of water at 20°C. How much energy is needed to heat it to 100°C? (c = 4200 J/kg°C)
- m = 1.0 kg, ΔT = 80°C. Q = 1.0 × 4200 × 80 = 4200 × 80.
- Compute: 4200 × 80 = 336,000 J.
- Answer: 336,000 J (336 kJ).
6. (Conduction) A metal rod requires 12,600 J to raise its temperature by 30°C. If the metal's specific heat is 420 J/kg°C, find the mass of the rod.
- Given Q = 12,600 J, c = 420 J/kg°C, ΔT = 30°C. Use Q = m c ΔT → m = Q / (c ΔT).
- Compute denominator: c ΔT = 420 × 30 = 12,600.
- m = 12,600 / 12,600 = 1.0 kg.
- Answer: 1.0 kg.
7. (Convection) A room has a heater that warms 2 kg of air (effective mass considered) from 20°C to 30°C. If specific heat of air ≈ 1000 J/kg°C (approx), how much heat is required?
- m = 2 kg, c ≈ 1000 J/kg°C, ΔT = 10°C.
- Q = m c ΔT = 2 × 1000 × 10 = 20,000 J.
- Answer: 20,000 J.
8. (Convection) A heated plate warms a thin layer of air of mass 0.05 kg by 15°C. Take c_air = 1000 J/kg°C. Find heat gained by the air.
- m = 0.05 kg, c = 1000 J/kg°C, ΔT = 15°C. Q = 0.05 × 1000 × 15.
- Compute: 0.05 × 1000 = 50; 50 × 15 = 750 J.
- Answer: 750 J.
9. (Convection — practical) A fan increases convective heat loss and causes a person to lose heat at a rate of 40 J/s. How much heat is lost in 5 minutes?
- Power = 40 J/s; time = 5 minutes = 5 × 60 = 300 s.
- Energy = power × time = 40 × 300 = 12,000 J.
- Answer: 12,000 J.
10. (Convection) 0.2 kg of air at 30°C is cooled to 20°C by a breeze. With c_air = 1000 J/kg°C, calculate heat removed.
- m = 0.2 kg, ΔT = 10°C. Q = 0.2 × 1000 × 10 = 2000 J.
- Answer: 2000 J.
11. (Radiation) A surface receives 500 J of solar energy. If a black surface absorbs 90% and a white surface absorbs 25%, how much energy does each absorb?
- Black absorbs 90%: 0.90 × 500 = 450 J.
- White absorbs 25%: 0.25 × 500 = 125 J.
- Answer: Black = 450 J; White = 125 J.
12. (Radiation) A shiny foil reflects 85% of incident radiation. If 2000 J fall on it, how much is reflected and how much absorbed?
- Reflected = 0.85 × 2000 = 1700 J. Absorbed = 2000 − 1700 = 300 J.
- Answer: Reflected 1700 J; Absorbed 300 J.
13. (Radiation) If sunlight delivers 1360 W/m² at top of atmosphere, estimate energy received by 0.5 m² in 10 minutes (use 1360 W/m² as incident power; 1 W = 1 J/s).
- Power on 0.5 m² = 1360 × 0.5 = 680 W (i.e., 680 J/s).
- Time = 10 minutes = 600 s. Energy = power × time = 680 × 600 = 408,000 J.
- Answer: 408,000 J.
14. (Radiation) A dark roof absorbs 70% of incident sunlight. If incident energy on the roof is 15,000 J during an hour, how much energy is absorbed?
- Absorbed = 0.70 × 15,000 = 10,500 J.
- Answer: 10,500 J.
15. (Water cycle — Latent heat) How much heat is required to evaporate 0.1 kg of water? (Use latent heat of vaporisation L_v = 2.26 × 10^6 J/kg)
- Q = m × L_v = 0.1 × 2.26×10^6 J = 0.1 × 2,260,000 J.
- Compute: 0.1 × 2,260,000 = 226,000 J.
- Answer: 226,000 J.
16. (Groundwater conduction) If surface heating raises soil temperature by 5°C to a depth where 2 kg of soil (effective) is warmed, and soil specific heat ≈ 800 J/kg°C, find heat conducted into that layer.
- m = 2 kg, c_soil ≈ 800 J/kg°C, ΔT = 5°C. Q = 2 × 800 × 5.
- Compute: 2 × 800 = 1600; 1600 × 5 = 8,000 J.
- Answer: 8,000 J conducted into that soil layer.
17. (Evaporation cooling) A wet cloth has 0.02 kg of water evaporated by breeze. Heat used = m L_v. Calculate cooling (use L_v = 2.26×10^6 J/kg).
- m = 0.02 kg. Q = 0.02 × 2.26×10^6 = 0.02 × 2,260,000.
- Compute: 2,260,000 × 0.02 = 45,200 J.
- Answer: 45,200 J of heat removed (cooling effect).
18. (Application) A 1000 W heater runs for 10 minutes to heat a closed space. How much energy is supplied? How many grams of water at 20°C could be heated to 30°C by that energy? (Assume all energy goes into heating water; c = 4200 J/kg°C.)
- Energy supplied = power × time = 1000 W × (10 × 60 s) = 1000 × 600 = 600,000 J.
- To heat water by 10°C, Q per kg = c × ΔT = 4200 × 10 = 42,000 J/kg.
- Mass heated = total energy / energy per kg = 600,000 / 42,000 = 14.2857... kg.
- In grams: 14.2857 kg = 14,285.7 g ≈ 14,286 g (rounded).
- Answer: Energy = 600,000 J; can heat ≈ 14.29 kg (≈14,286 g) of water by 10°C.
19. (Soil vs water heating) 1 kg of dry soil (c=800 J/kg°C) and 1 kg of water (c=4200) are heated with 84,000 J. Find temperature rise for each.
- For soil: ΔT_soil = Q / (m c) = 84,000 / (1 × 800) = 84,000 / 800 = 105°C.
- For water: ΔT_water = 84,000 / (1 × 4200) = 84,000 / 4200 = 20°C.
- Answer: Soil temperature rises by 105°C (shows low heat capacity), water rises by 20°C (high heat capacity).
20. (Practical) A solar panel of area 2 m² receives average solar power 500 W/m² for 3 hours. If panel converts 15% to electrical energy, calculate electrical energy produced.
- Incident power on panel = 500 × 2 = 1000 W.
- Time = 3 hours = 3 × 3600 = 10,800 s.
- Total incident energy = power × time = 1000 × 10,800 = 10,800,000 J.
- Electrical energy = 15% of that = 0.15 × 10,800,000 = 1,620,000 J.
- Answer: 1,620,000 J (or 1.62 MJ) of electrical energy.