Question
A heat-vs-temperature graph for 1 kg of ice at −20°C being heated to steam at 120°C shows several segments. Identify the segments and calculate the total heat required. Use cice=2100 J/(kg⋅K), Lf=334000 J/kg, cw=4186 J/(kg⋅K), Lv=2260000 J/kg, csteam=2010 J/(kg⋅K).
Solution — Step by Step
Reading the graph from left to right:
- Ice from −20°C to 0°C (sloped — temperature rises)
- Ice melting at 0°C (flat — phase change)
- Water from 0°C to 100°C (sloped)
- Water boiling at 100°C (flat — phase change)
- Steam from 100°C to 120°C (sloped)
Q1=mciceΔT=1×2100×20=42000 J
Q2=mLf=1×334000=334000 J
Q3=mcwΔT=1×4186×100=418600 J
Q4=mLv=1×2260000=2260000 J
Q5=mcsteamΔT=1×2010×20=40200 J
Qtotal=42000+334000+418600+2260000+40200=3094800 J
Final answer: Qtotal≈3.09×106 J≈3.09 MJ.
Why This Works
Heat input does one of two things: raise temperature (Q=mcΔT) or change phase (Q=mL). On the heat-vs-temperature graph, sloped segments are temperature changes; flat plateaus are phase changes (heat goes in but temperature stays put).
Latent heat of vaporisation (2260 kJ/kg) dominates — it’s about 5 times bigger than the latent heat of fusion. That’s why steam burns are far more dangerous than hot-water burns: condensing steam dumps a huge amount of energy on contact.
Alternative Method
Group similar segments first. Sensible heat (slopes): 42+418.6+40.2=500.8 kJ. Latent heat (plateaus): 334+2260=2594 kJ. Total ≈3094.8 kJ. Same answer, helps catch arithmetic errors.
JEE Main loves to ask which segment requires the most heat. For water, vaporisation always wins by a wide margin. For other substances, check the latent heat values — some metals have larger fusion than vaporisation latent heats per mole, but per kg, vaporisation usually dominates.
Common Mistake
Forgetting one of the five segments — usually the steam-heating segment after 100°C. Always count: ice phase, fusion plateau, water phase, vaporisation plateau, steam phase. Five segments for ice → superheated steam.