Find the amount of heat required to convert one gram of ice at -10°C into steam at 100°C

Heat required to convert 1g ice at -10°C to steam at 100°C

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Find the amount of heat required to convert one gram of ice at -10°C into steam at 100°C — this classic calorimetry problem is solved by adding the heat for warming ice to 0°C, melting, heating water to 100°C and vaporisation.

Given data & Focus Keyword

• Mass, m = 1 g = 0.001 kg
• Initial temperature of ice = −10°C
• Final temperature (steam) = 100°C
• Focus Keyword: Find the amount of heat required to convert one gram of ice at -10°C into steam at 100°C

Constants / Values used

• Specific heat of ice, c_ice ≈ 2100 J·kg⁻¹·K⁻¹
• Specific heat of water, c_water ≈ 4200 J·kg⁻¹·K⁻¹
• Latent heat of fusion of ice, L_f ≈ 3.36 × 10⁵ J·kg⁻¹
• Latent heat of vaporization of water, L_v ≈ 2.26 × 10⁶ J·kg⁻¹

Solution — Step by step

1. Heat to raise ice from −10°C to 0°C:
   Q₁ = m · c_ice · ΔT = 0.001 × 2100 × 10 = 21 J
2. Heat to melt ice at 0°C:
   Q₂ = m · L_f = 0.001 × 3.36×10⁵ = 336 J
3. Heat to raise water from 0°C to 100°C:
   Q₃ = m · c_water · ΔT = 0.001 × 4200 × 100 = 420 J
4. Heat to vaporise water at 100°C:
   Q₄ = m · L_v = 0.001 × 2.26×10⁶ = 2260 J

Total heat required = Q₁ + Q₂ + Q₃ + Q₄ = 21 + 336 + 420 + 2260 = ≈ 3037 J (≈ 3.04 kJ).

Heat required to convert 1g ice at -10°C to steam at 100°C

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