Question

Oblicz ile ciepła trzeba dostarczyć aby 5 kg lodu o tem. -20°C zamienić w parę o tem. 120°C

Answer 1

Verified answer

m = 5 kg

t₁ = -20°C

t₂ = 0°C

t₃ = 100°C

t₄ = 120°C

cl = 2100 J/kg*C

ct = 334000 J/kg

cw = 4200 J/kg*C

cp = 2300000 J/kg

1. Q₁ -> ogrzanie lodu do 0°C :

Q₁ = m×cl×(t₂ - t₁)

Q₁ = 5 × 2100 × (0 - 20) = -210000 [J]

2. Q₂ -> zamiana lodu w wodę :

Q₂ = m×ct

Q₂ = 5 × 334000 = 1670000 [J]

3. Q₃ -> ogrzanie wody do 100°C :

Q₃ = m×cw×(t₃ - t₂)

Q₃ = 5 × 4200 × (100 - 0) = 2100000 [J]

4. Q₄ -> zamiana wody w parę :

Q₄ = m×cp

Q₄ = 5 × 2300000 = 11500000 [J]

5. Q₅ -> ogrzanie pary do 120°C

Q₅ = m×cp×(t₄ - t₃)

Q₅ = 5 × 2300000 × (120 - 100) = 230000000 [J]

Całkowite ciepło :

Q = Q₁ + Q₂+ Q₃ + Q₄ + Q₅

Q = -210000 + 1670000 + 2100000 + 11500000 + 230000000 = 245060000 [J]

Q = 245,06 MJ