Question

What quantity of heat is needed to convert 5 kg of ice at -19 degrees C to steam at 100 degrees C?

Heat = 13590000 Joules -wrong

Answer 1

Mass of ice = m = 5 kg

Initial temperature of ice = T₁ = -19 ^oC

Melting point of ice = T₂ = 0 ^oC

Boiling point of water = T₃ = 100 ^oC

Specific heat of ice = C₁ = 2100 J/(kg.^oC)

Specific heat of water = C₂ = 4186 J/(kg.^oC)

Latent heat of fusion of water = L₁ = 3.34 x 10⁵ J/kg

Latent heat of vaporization of water = L₂ = 2.26 x 10⁶ J/kg

Heat needed to convert the ice from -19^oC to steam at 100^oC = H

H = mC₁(T₂ - T₁) + mL₁ + mC₂(T₃ - T₂) + mL₂

H = (5)(2100)(0 - (-19)) + (5)(3.34x10⁵) + (5)(4186)(100 - 0) + (5)(2.26x10⁶)

H = 1.526 x 10⁷ J

Heat needed to convert 5 kg of ice at -19^oC to steam at 100^oC = 1.526 x 10⁷ J