ii) Compute the total heat required to convert 0.125 Kg of ice at –125°C to steam at 125 °C. (Cice = 2090 J/kg.°C, Cwater = 4186 J/kg.°C, Csteam = 2010 J/kg.°C, %3D Lf = 3.33 × 10$ J/kg, Lv = 2.26 × 10° J/kg.)

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Chapter14: Heat And Heat Transfer Methods
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ii) Compute the total heat required to convert 0.125 Kg of ice at –125°C to steam at 125 °C.
(Cice = 2090 J/kg.°C, Cwater = 4186 J/kg.°C, Csteam = 2010 J/kg.°C,
Lf = 3.33 × 10$ J/kg, L, = 2.26 × 106 J/kg.)
Transcribed Image Text:ii) Compute the total heat required to convert 0.125 Kg of ice at –125°C to steam at 125 °C. (Cice = 2090 J/kg.°C, Cwater = 4186 J/kg.°C, Csteam = 2010 J/kg.°C, Lf = 3.33 × 10$ J/kg, L, = 2.26 × 106 J/kg.)
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