The tank, cylindrical in shape has a radius 1.0m and maximum depth of water 2.0m

ID: 809053 • Letter: T

Question

The tank, cylindrical in shape has a radius 1.0m and maximum depth of water 2.0m.

The tank is filled with water at 25oC which must be brought to boil at 108oC due to pressurization.

If the current to the heaters is 110kW, how long will it take a full tank of water to be heated to 108oC?

The specific heat capacity of water c=4190J/kg oC

so using Q=mcT i thought because Q=W/t and W=110kW I could work out time.

and work out m=pV and using density of water=1000kg/m^3 and V=pi x 1^2 x 2=2pi m^3 Thus m=6283 kg.

Substituing these values back into the equation gives me an answer something totally different to the actual answer.

The actual answer is 6.5 hours. Could someone please explain why?

above question can not be answered directly in a single step as we can't apply specific heat of liquid water at different phases.

water = liquid (T<=100'C) specific heat C= 4190 J/KgK

Q= mCdT

Q1= 6283*4190(75)

= 1974432750 J

water (T= 100'C phase conversion)

Q = mL (latent heat of vaporisation = 2262000J/KgK)

6283*2262000

= 114212146000J

water(T>100'C)

Q = mCdT (C = 2000k/KgK )

Q3= 6283*2000*8

= 100528000

so total heat = 16287106750J = 110*1000(t)

t= 41.12 Hours

phase conversion(at 100'C from liquid to vapour)

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