Asked by JOHN
A pot containing 1.09 liter of water, initially at 18°C, is placed on a 1101 W electric heating element.
(a) How much heat must be supplied to the water to bring it to a boil?
(b) How much heat is necessary to boil all the water away?
What is the minimum time required to boil all the water away?
(a) How much heat must be supplied to the water to bring it to a boil?
(b) How much heat is necessary to boil all the water away?
What is the minimum time required to boil all the water away?
Answers
Answered by
Elena
L = 2260000 J/kg is the heat of vaporization,
ρ =1000 kg/m^3 is the density of the water,
c = 4185.5 J/(kg•K) is heat capacity of water.
Q1 = c•m•ΔT= c•m•ΔT = c•ρ•V •ΔT = =4180•1000•1.09•10^-3•82 = 3.73•10^5 J
Q2 = L•m = L• ρ•V =2260000 • 1000•1.09•10^-3 =2.46•10^6 J
Q = Q1+Q2 = P•t,
t = (Q1+Q2)/P =
= 3.73•10^5+2.46•10^6)/ 1101= 2574 s = =42.89 min.
ρ =1000 kg/m^3 is the density of the water,
c = 4185.5 J/(kg•K) is heat capacity of water.
Q1 = c•m•ΔT= c•m•ΔT = c•ρ•V •ΔT = =4180•1000•1.09•10^-3•82 = 3.73•10^5 J
Q2 = L•m = L• ρ•V =2260000 • 1000•1.09•10^-3 =2.46•10^6 J
Q = Q1+Q2 = P•t,
t = (Q1+Q2)/P =
= 3.73•10^5+2.46•10^6)/ 1101= 2574 s = =42.89 min.
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