Question

in cit timp un fierbator de apa cu puterea de 1kw incalzeste 10 l de apa de la 20 grade celsius pina la 80 grade celsius . pierderile de energie constituie 15 % . ( c = 4,19 kj/(kg*grade celsius )apa ) q apa =1000 kg/m cubi

Answer 1

Se da:
P=1kW=1000W
V=10l=0,01m³
t1=20°C
t2=80°C
pe=15%⇒n=85%=0,85
c=4,19kJ/kg°C=4190J/kg°C
p=1000kg/m³
t=?s

Formule:
n=Qu/Lt
n=cmΔt/Pt
n=cVp(t2-t1)/Pt
t=cVp(t2-t1)/Pn

Calcule:
t=4190*0,01*1000*(80-20)/1000*0,85≈2958s