[tex]dane:\\T=0^{o}C\\P = 800 \ W = 0,8 \ kW\\t = 3 \ h= 3\cdot3600 \ s = 10 \ 800 \ s\\c_{t} = 334 \ 000\frac{J}{kg}\\szukane:\\m = ?\\W = ?[/tex]
Rozwiązanie
Obliczmam masę lodu m:
[tex]P = \frac{W}{t} \ \ /\cdot t\\\\W = P\cdot t\\oraz\\W = Q = c_{t}\cdot m\\\\c_{t}\cdot m = P\cdot t \ \ /:c_{t}\\\\m = \frac{P\cdot t}{c_{t}}\\\\m = \frac{800 \ W\cdot10800 \ s}{334000\frac{J}{kg}}= \frac{800\frac{J}{s}\cdot10800 \ s}{334000\frac{J}{kg}}\\\\\boxed{m = 25,868 \ kg \approx25,9 \ kg}[/tex]
Obliczam zużycie energii elektrycznej :
[tex]E_{el} = W = P\cdot t\\\\W = 0,8 \ kW\cdot3 \ h\\\\\boxed{W = 2,4 \ kWh}[/tex]
