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a) What minimum heat is needed

to bring 250 g of water at 40 ∘ C to the boiling point and completely boil it away? The specific heat of water is C= 4190 J/(kg⋅C)and its heat of vaporization is  LV 22.6×105 J/C

b) How much ice at 0 ∘ C is needed to make rain out of the water vapors of part a and bring it back to the same temperature of 40 ∘ C ?

The heat of meting for ice is Lm  = 334 J/kg.

We consider the system is completely isolated from the environment.

Show your complete solution.

Q = mC ( T2 - T1)

Q = m LV

Q = m Lm

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m
water - 250 g = 0.230 kg
T : 40 c
T 2 = 100 0 ( ( boiling point )
C = 4190 3 / rgic
L = 22.6 x 103 5 / kg
( a )
2 = mCAT - mLy
Q : (0.25 Kg) ( 4190 5 ) (100 0 - 40 c) +(0.25 kg)(22.6x10
2: 628505...

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