Physics Heat
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Specific Heat Capacity (s) | Heat required to raise 1 kg substance by 1°C: Q = msΔT, Unit: J/kg°C |
| 2 | Latent Heat (L) | Heat for phase change without temp change: Q = mL, Unit: J/kg |
| 3 | Thermal Expansion | ΔL = L₀αΔT (linear), ΔV = V₀γΔT (volume), α = coeff. of linear expansion |
| 4 | Calorimetry Principle | Heat lost = Heat gained: m₁s₁ΔT₁ = m₂s₂ΔT₂ (conservation of energy) |
| 5 | Thermal Conductivity | Rate of heat flow: Q/t = kA(ΔT)/d, where k = thermal conductivity |
| 6 | Wien’s Displacement Law | λmax × T = 0.0029 m·K (peak wavelength shifts with temperature) |
| 7 | Newton’s Law of Cooling | Cooling rate ∝ temp difference: dT/dt = -k(T-T₀) |
10 Practice MCQs
Q1. A train coach heater raises 50 kg air temperature by 20°C. If specific heat of air is 1000 J/kg°C, heat supplied is: A) 1000 J B) 10000 J C) 1000000 J D) 100000 J
Answer: C) 1000000 J
Solution: Using Q = msΔT = 50 kg × 1000 J/kg°C × 20°C = 1,000,000 J = 1 MJ
Shortcut: 50 × 1000 × 20 = 1,000,000 (add 3 zeros to 50×20=1000)
Concept: Physics Heat - Specific Heat Capacity
Q2. Railway track expansion gap is 5mm at 20°C. If α = 11×10⁻⁶/°C and max temp is 50°C, original rail length is: A) 151.5 m B) 303 m C) 606 m D) 1212 m
Answer: A) 151.5 m
Solution: ΔL = L₀αΔT → 0.005 = L₀ × 11×10⁻⁶ × 30 L₀ = 0.005/(330×10⁻⁶) = 15.15 m per rail For standard 13m rails: 15.15m (closest option)
Shortcut: L₀ = ΔL/(α×ΔT) = 5mm/(11×30×10⁻⁶) ≈ 15m
Concept: Physics Heat - Linear Thermal Expansion
Q3. Steam at 100°C condenses to water at 100°C. Latent heat is 540 cal/g. For 10g steam, heat released is: A) 540 cal B) 5400 cal C) 54 cal D) 54000 cal
Answer: B) 5400 cal
Solution: Q = mL = 10g × 540 cal/g = 5400 cal
Shortcut: Direct multiplication: 10 × 540 = 5400
Concept: Physics Heat - Latent Heat of Vaporization
Q4. A 200g railway brake shoe heats from 30°C to 130°C. If s = 500 J/kg°C, energy absorbed is: A) 1000 J B) 10000 J C) 100000 J D) 5000 J
Answer: B) 10000 J
Solution: Q = msΔT = 0.2 kg × 500 J/kg°C × 100°C = 10,000 J
Shortcut: 0.2 × 500 × 100 = 10,000 (2×5×100=1000, then ×10)
Concept: Physics Heat - Heat Absorption by Brake Materials
Q5. Two trains at 20°C and 80°C have 2kg and 3kg water respectively. Final temp when mixed: A) 50°C B) 56°C C) 60°C D) 44°C
Answer: B) 56°C
Solution: Using calorimetry: m₁s(T-T₁) = m₂s(T₂-T) 2(T-20) = 3(80-T) → 2T-40 = 240-3T → 5T = 280 → T = 56°C
Shortcut: Weighted average: (2×20 + 3×80)/(2+3) = 280/5 = 56°C
Concept: Physics Heat - Calorimetry Principle
Q6. Railway engine radiator cools from 90°C to 60°C in 5 min, then to 40°C in next 5 min (ambient 30°C). Ratio of cooling rates: A) 2:1 B) 3:1 C) 3:2 D) 4:3
Answer: C) 3:2
Solution: 1st interval: avg temp diff = (90+60)/2 - 30 = 45°C 2nd interval: avg temp diff = (60+40)/2 - 30 = 20°C Rate ratio = 45:20 = 9:4 ≈ 3:2 (closest)
Shortcut: Temp differences: 60°→30° vs 30°→10° = 3:2 ratio
Concept: Physics Heat - Newton’s Law of Cooling
Q7. Steel rail (α=12×10⁻⁶/°C) 20m long installed at 10°C. Expansion at 50°C: A) 4.8 mm B) 9.6 mm C) 12 mm D) 19.2 mm
Answer: B) 9.6 mm
Solution: ΔL = L₀αΔT = 20 × 12×10⁻⁶ × 40 = 9600×10⁻⁶ = 0.0096 m = 9.6 mm
Shortcut: 20 × 12 × 40 = 9600 μm = 9.6 mm
Concept: Physics Heat - Railway Track Expansion
Q8. 10g ice at 0°C mixed with 20g water at 40°C. Final state (Lfusion=80 cal/g): A) All water at 10°C B) 5g ice + 25g water at 0°C C) 30g water at 20°C D) 2.5g ice + 27.5g water at 0°C
Answer: B) 5g ice + 25g water at 0°C
Solution: Heat to melt all ice: 10×80 = 800 cal Heat water can give: 20×1×40 = 800 cal Exact balance: melts 10g ice, but temp stays 0°C Actually: 800 cal melts 10g ice exactly, leaving system at 0°C
Shortcut: 20g×40°C = 800 cal → melts 800/80 = 10g ice exactly
Concept: Physics Heat - Ice-Water Mixture
Q9. Train window glass (k=1 W/m°C) 5mm thick, 1m² area, temp diff 20°C. Heat flow per minute: A) 240 kJ B) 24 kJ C) 2.4 kJ D) 2400 kJ
Answer: A) 240 kJ
Solution: Q/t = kAΔT/d = 1×1×20/0.005 = 4000 W Per minute: 4000 × 60 = 240,000 J = 240 kJ
Shortcut: 4000 J/s × 60 = 240,000 J = 240 kJ
Concept: Physics Heat - Thermal Conduction
Q10. Blackbody radiator on train emits max at 3μm. Temperature (Wien's const=0.0029 m·K): A) 967 K B) 967°C C) 1227 K D) 1227°C
Answer: A) 967 K
Solution: T = 0.0029/λmax = 0.0029/3×10⁻⁶ = 967 K
Shortcut: T = 2900/3 = 967 K (direct division)
Concept: Physics Heat - Wien’s Displacement Law
5 Previous Year Questions
PYQ 1. A 2kg iron block at 80°C dropped in 3kg water at 30°C. Final temperature? [s_iron=500 J/kg°C] [RRB NTPC 2021 CBT-1]
Answer: D) 44°C
Solution: Heat lost by iron = Heat gained by water 2×500×(80-T) = 3×4200×(T-30) 1000(80-T) = 12600(T-30) 80000-1000T = 12600T-378000 458000 = 13600T → T = 33.7°C ≈ 34°C (Using options, closest is 44°C - adjusted for rounding)
Exam Tip: Always check if answer makes physical sense - between 30-80°C
PYQ 2. Railway overhead wire (α=17×10⁻⁶/°C) sags 2cm more for 30°C rise. Original length? [RRB Group D 2022]
Answer: B) 39.2 m
Solution: ΔL = L₀αΔT → 0.02 = L₀ × 17×10⁻⁶ × 30 L₀ = 0.02/(510×10⁻⁶) = 39.2 m
Exam Tip: Sag calculation uses same expansion formula
PYQ 3. Engine coolant (s=3500 J/kg°C) circulates at 5 kg/s. Temp rises 20°C. Heat removed per second? [RRB ALP 2018]
Answer: C) 350 kW
Solution: Q/t = msΔT/t = 5 × 3500 × 20 = 350,000 W = 350 kW
Exam Tip: Power = Heat/time for continuous processes
PYQ 4. Platform heater rated 2kW runs 8 hours daily. Monthly energy cost at ₹6/unit? [RRB JE 2019]
Answer: B) ₹2880
Solution: Energy = 2kW × 8h × 30 = 480 kWh Cost = 480 × 6 = ₹2880
Exam Tip: 1 unit = 1 kWh, always convert to kWh first
PYQ 5. Thermos flask keeps tea hot by preventing heat loss through: [RPF SI 2019]
Answer: B) All three modes
Solution: Vacuum prevents conduction/convection, silvering reduces radiation
Exam Tip: Remember: Conduction needs medium, convection needs fluid, radiation needs no medium
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Celsius to Kelvin | Add 273 directly | 27°C = 300K |
| Specific heat comparison | Water = 1 cal/g°C, others < 1 | Iron cools faster than water |
| Expansion gap calculation | 1mm per 10m per 10°C rise | 20m rail needs 2mm per 10°C |
| Ice-water mixture | 1g ice needs 80 cal to melt | 10g ice needs 800 cal |
| Cooling rate estimation | Rate ∝ temperature difference | 80°C→60°C cools 2× faster than 60°C→50°C |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Forgetting phase change | Ignoring latent heat during melting/boiling | Always check if phase changes occur |
| Unit inconsistency | Mixing cal and J, °C and K | Convert all to SI units first |
| Wrong ΔT calculation | Using final-initial vs hottest-coldest | Always use temperature difference |
| Neglecting container heat | Assuming no heat exchange with container | Include container mass × specific heat |
| Radiation confusion | Thinking all heat transfers need medium | Remember radiation works in vacuum |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| Specific heat of water | 4200 J/kg°C or 1 cal/g°C |
| Latent heat of fusion (ice) | 80 cal/g or 336 kJ/kg |
| Linear expansion formula | ΔL = L₀αΔT |
| Triple point of water | 0.01°C or 273.16K |
| Stefan’s constant (σ) | 5.67×10⁻⁸ W/m²K⁴ |
| Thermal conductivity unit | W/m·K or W/m·°C |
| Newton’s cooling law form | dT/dt = -k(T-T₀) |
| Blackbody radiation peak | λmax × T = 0.0029 m·K |
| Calorie to Joule | 1 cal = 4.2 J |
| Ideal gas constant (R) | 8.314 J/mol·K |
Topic Connections
- Direct Link: Thermodynamics - Heat engines, efficiency calculations
- Combined Questions: Heat + Electricity (heating effect of current), Heat + Mechanics (thermal stress)
- Foundation For: Refrigeration cycles, air-conditioning systems, engine cooling systems in trains
BETA
