An electric kettle takes 10 mins to bring water from 30 degree C to 100. Find the extra time required to vaporize half of the boiling water in the kettle, assume that no heat is lost to the surroundings.
specific heat capacity of water = 4200 J kg-1 C-1
specific latent heat of v. of water = 2.268 x 10^6 J kg-1
Phy very hard question
2008-09-04 3:31 am
回答 (2)
2008-09-04 3:57 am
✔ 最佳答案
Let the electric power of the kettle be P and the mass of water be mE=mc∆T
Pt=mc∆T
P(10)(60)=m(4200)(100-30)
P=490m
The extra energy to vaporize half of the boiling water
=(m/2)lv
=(m/2)(2.268 x 10^6)
BY E=Pt
(m/2)(2.268 x 10^6)=490m(t)
t~2314s
t~39min
2008-09-04 4:01 am
Heat absorbed by water rising from temperature of 30 deg. to 100 deg
= m.(4200)(100-30) J = 294000m J
[where m is the mass of water]
Hence, power of the electric kettle = 294000m/10 J/min = 29400m J/min
Heat required to vapourized water of mass (m/2)
= (m/2)(2.268x10^6) J
Since heat supplied by the kettle = 29400m.t J
where t is the time in minutes required to vapourize water of mass [m/2]
Therefore, we have,
29400m.t = (m/2)(2.268x10^6)
i.e. t = 2.286^6/(2x29400) minutes
= m.(4200)(100-30) J = 294000m J
[where m is the mass of water]
Hence, power of the electric kettle = 294000m/10 J/min = 29400m J/min
Heat required to vapourized water of mass (m/2)
= (m/2)(2.268x10^6) J
Since heat supplied by the kettle = 29400m.t J
where t is the time in minutes required to vapourize water of mass [m/2]
Therefore, we have,
29400m.t = (m/2)(2.268x10^6)
i.e. t = 2.286^6/(2x29400) minutes
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