Question

Construct the flow diagram and estimate the total heat required to convert 550 g ice at –75o C to steam at 1850 C.

(Cice = 2090 J/kg.°C, Cwater = 4186 J/kg.°C, Csteam = 2010 J/kg.°C, Lf = 3.33 × 105 J/kg, Lv = 2.26 × 106 J/kg.)

Answer 1

Given mass of ice m = 550 g = 0.55 kg

Ci =2090 J/kg.°C,

Cw = 4186 J/kg.°C,

Cs = 2010 J/kg.°C,

Lf = 3.33 × 105 J/kg,

Lv = 2.26 × 106 J/kg

Heat required -75 oC to 0oC ice temperature = mCiT

= 0.55*2090*(0-(75))

= 0.55*2090*75

= 86,212.5‬ J

Heat required 0 oC ice to 0oC water temperature = mLf

= 0.55* 3.33 × 105

= 18315‬0 J

Heat required 0 oC water to 100oC water temperature = mCwT

= 0.55*4186*(100-0)

= 0.55*4186*100

= 230230 J

Heat required 100 oC water to 100oC steam temperature = mLv

= 0.55*2.26*106

=1243000 J

(If final temperature of stream T = 185 oC )

Then

Heat required 100 oC steam to 185 oC steam temperature = mCsT

= 0.55* 2010 *(185 -100)

= 0.55*2010*85

= 93,967.5‬ J

total heat require = + + + +

= 86,212.5 +  18315‬0 + 230230 + 1243000 + 93,967.5

= 1,836,560‬ J .........................Ans

Note (If final temperature of stream T = 1850 oC )

Then

Heat required 100 oC steam to 1850 oC steam temperature = mCsT

= 0.55* 2010 *(1850-100)

= 0.55*2010*1750

= 1,934,625‬ J

total heat require = + + + +

= 86,212.5 +  18315‬0 + 230230 + 1243000 + 1,934,625

= 3,677,217.5 J .........................Ans