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Q.Calculate the amount of heat required to convert 1 kg of ice at –10°C to steam at 100°C at normal pressure.
Given: Specific heat of ice = 2100 J/kg-K, Latent heat of fusion of ice = 3.36 × 10⁵ J/kg, Specific heat of water = 4200 J/kg-K, Latent heat of steam = 2.25 × 10⁶ J/kg.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2026Subjective· 3mImportance★★★★★
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Sum the heat for 4 stages — warm ice, melt it, warm the water, vaporize it — to get about 3.03×1063.03\times10^6 J.

Converting 1 kg of ice at −10∘C-10^\circ\text{C} all the way to steam at 100∘C100^\circ\text{C} requires four separate stages, each needing its own heat input:

Stage 1 — warm the ice from −10∘C-10^\circ\text{C} to 0∘C0^\circ\text{C}:

Q1=mciceΔT=1×2100×10=21,000 JQ_1 = mc_{\text{ice}}\Delta T = 1\times2100\times10 = 21{,}000\text{ J}

Stage 2 — melt the ice at 0∘C0^\circ\text{C} (phase change, no temperature rise):

Q2=mLf=1×3.36×105=336,000 JQ_2 = mL_f = 1\times3.36\times10^5 = 336{,}000\text{ J}

Stage 3 — warm the water from 0∘C0^\circ\text{C} to 100∘C100^\circ\text{C}:

Q3=mcwaterΔT=1×4200×100=420,000 JQ_3 = mc_{\text{water}}\Delta T = 1\times4200\times100 = 420{,}000\text{ J}

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