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T = 0 ⁰ C
ΔT =1 ⁰ C
Rozw.:
Aby go stopić potrzeba:
Q₁ = m × Ct = m[kg] × 335000 [J/kg]
Aby go ogrzać o 1 ⁰ C:
Q₂ = m × ΔT × Cw = m [kg] × 1 ⁰ C × 4200 [J/kg ⁰ C] = m[kg] × 4200 [J/kg]
Q₁/Q₂ = m [kg]× 335000 [J/kg] / m[kg] × 4200 [J/kg] = 79,76