What Is the Amount of Heat Required to Convert 1 kg of Ice at –10 °C to Steam at 100 °C?
What is the amount of heat required to convert 1 kg of ice at –10 °C to steam at 100 °C? This is a standard calorimetry problem involving heating, melting, heating water, and vaporization steps.
Steps Involved in Heat Calculation
The process includes four stages:
- Heating ice from –10 °C to 0 °C
- Melting ice at 0 °C to water at 0 °C
- Heating water from 0 °C to 100 °C
- Converting water at 100 °C to steam at 100 °C
Heat Required in Each Stage
Commonly used values for water/ice: specific heat of ice ≈ 2100 J/kg·K, of water ≈ 4200 J/kg·K; latent heat of fusion ≈ 3.36×10⁵ J/kg, latent heat of vaporization ≈ 2.26×10⁶ J/kg. :contentReference[oaicite:0]{index=0}
- Heating ice from –10 °C to 0 °C: Q₁ = m·cice·ΔT = 1×2100×(0 – (–10)) = 21 000 J
- Melting ice at 0 °C: Q₂ = m·Lfusion = 1×3.36×10⁵ = 336 000 J
- Heating water from 0 °C to 100 °C: Q₃ = m·cwater·ΔT = 1×4200×100 = 420 000 J
- Vaporizing water at 100 °C to steam: Q₄ = m·Lvaporization = 1×2.26×10⁶ = 2 260 000 J
Total Heat Required (Qtotal)
Qtotal = Q₁ + Q₂ + Q₃ + Q₄ ≈ 21 000 + 336 000 + 420 000 + 2 260 000 = ≈ 3.037 × 10⁶ J (≈ 3.04 MJ)
Note: This calculation assumes ideal conditions — no heat loss to surroundings, no pressure deviations, and standard values for specific and latent heats. Actual energy needed may vary in real conditions.
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Keywords
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