9) How much energy in Kcal is needed to take 340 g of water at 6.0...

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Answer 1

Answer #1

ANSWER

The process involve three steps (i) heating water from 60 ^oC to 100 ^oC. (ii) Vaporization (iii) heating steam from 100 ^oC to 127 ^oC.

Thus total heat required (Q) is given as

Q = Heat required in Process (i) + Heat required in Process (ii) + Heat required in Process (iii)

Q = ( m X c X ΔT )_w + Heat of vaporization + ( m X c X ΔT )_v

( m X c X ΔT )_w = product of mass , specific heat and temperature difference of water

( m X c X ΔT )_v = product of mass , specific heat and temperature difference of vapor

Q = [340g X1.0 cal/g^oC X(100^oC- 60^oC)]_w + (340g X 540Cal/g) +

[340gX0.48 cal/g^oC X (127^oC - 100^oC)]_w

Q = 13600 + 183600 + 4406.8 = 201.6KCal

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Answer 2

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