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m = 300 g = 0,3 kg
T₁ = - 5⁰C
temperatura topnienia Tt = 0⁰C
T₂ = 5 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 - (-5⁰C) = 5⁰C
Q₁ = 0,3 kg × 2100 [J/kg×⁰C] × 5⁰C = 3150 J
2. Ciepło potrzebne do stopienia lodu
Q₂ = m × Ct = 0,3 kg × 335000 [J/kg] = 100500J
3. Ciepło potrzebne do podgrzania do 5 C wody powstałej z lodu
Q₃ = m × Cw × ΔT
ΔT= T₂ - Tt = 5 ⁰C - 0⁰C = 5⁰C
Q₃ = 0,3kg × 4200 [J/kg×⁰C] × 5⁰C = 6300 J
Qc = Q₁ + Q₂ + Q₃ = 109950J