Electric cooker

During which time t does an electric cooker with power input P = 500 W and with efficiency n = 75% heat water with mass m = 2 kg and temperature t1 = 10°C to the boiling point (t2 = 100°C).
The specific heat capacity of water is c = 4 180 J. Kg-1. K-1

Correct result:

t =  33.44 min

Solution:

P=500 W n=75%=75100=0.75 m=2 kg t1=10t2=100=2006.4c=4180 J/kg/K  T=t2t1=2006.410=90 K Q=m c T=2 4180 90=752400 J  P2=P n=500 0.75=375 W  Q=P2 t  t2=QP2=752400375=100325=200625=2006.4 s  t=t2 min=t2/60  min=2006.4/60  min=33.44 min=83625 min



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