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m = 5 g = 0,005 kg
T₁ = - 10⁰C
temperatura topnienia Tt = 0⁰C
T₂ = 10 ⁰C
Cw(lodu) = 2100 [J/kg×⁰C]
Ct = 335000 [J/kg]
Cw(wody) ≈ 4200 [J/kg×⁰C] (
Rozw.:
1. Ciepło pobrane podczas ogrzania lodu do temp 0⁰C
Q₁ = m × Cw × ΔT
ΔT= Tt - T₁ = 0 ⁰C - (-10⁰C) = 10⁰C
Q₁ = 0,005 kg × 2100 [J/kg×⁰C] × 10⁰C = 105J
2. Ciepło potrzebne do stopienia lodu
Q₂ = m × Ct = 0,005 kg × 335000 [J/kg] = 1675J
3. Ciepło potrzebne do podgrzania do 10 C wody powstałej z lodu
Q₃ = m × Cw × ΔT
ΔT= T₂ - Tt = 10 ⁰C - 0⁰C = 10⁰C
Q₃ = 0,005kg × 4200 [J/kg×⁰C] × 10⁰C = 210J
Qc = Q₁ + Q₂ +Q₃ = 1990J