Asked by Anonymous
what should be the resistance of a heating coil which will be used to raise the temperature of 500 grams of water from 28 degrees celcius to 100 degrees celcius in 2 minutes, assuming that 25% of the heat is lost? the heater utilizes 110 volts.
Answers
Answered by
Allan Bernard Dela Vega
Boiling point=100C
Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)2/R
R=7.23ohms
Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)2/R
R=7.23ohms
Answered by
Allan Bernard Dela Vega
Boiling point=100C
Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)^2/R
R=7.23ohms
Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)^2/R
R=7.23ohms
Answered by
Wole
When can the final temperature be added to the initial temperature
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