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m = 300 g = 0,3 kg
T₁ = 0⁰C
temperatura topnienia Tt = 0⁰C
T₂ = 15 ⁰C
Ct = 335000 [J/kg]
Cw(wody) = 4200 [J/kg×⁰C]
Rozw.:
Ciepło potrzebne do stopienia lodu
Q₁ = m × Ct = 0,3 kg × 335000 [J/kg] = 100500J
Ciepło potrzebne do podgrzania do 15 C wody powstałej z lodu
Q₂ = m × Cw × ΔT
ΔT= T₂ - T₁ = 15 ⁰C - 0⁰C = 15⁰C
Q₂ = 0,3kg × 4200 [J/kg×⁰C] × 15⁰C = 18900J
Qc = Q₁ +Q₂ = 119400 J