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How much heat is required to evaporate 230 lb of water per hour under a pressure of 111.3 psig if the water enters the boiler at 190°F?

Answer :

Heat required to evaporate the water to steam is 5342 KJ

We know that,

[tex]Q_{t}[/tex] = c * m * ( [tex]T_{f}[/tex] - [tex]T_{i}[/tex] )

where,

[tex]Q_{t}[/tex] = Heat required to raise the temperature

c = Specific heat capacity

m = Mass

[tex]T_{f}[/tex] = Final temperature

[tex]T_{i}[/tex] = Initial temperature

Given that,

m = 230 lb = 104.33 kg

[tex]T_{i}[/tex] = 190 F = 87.78 °C

[tex]T_{f}[/tex] = 100 °C (Because water boils at 100 °C to turn into steam)

c = 4190 J / kg °C

[tex]Q_{t}[/tex] = c * m * ( [tex]T_{f}[/tex] - [tex]T_{i}[/tex] )

= 4190 * 104.33 * 12.22

[tex]Q_{t}[/tex] = 5341883.79 J = 5342 KJ

Heat is the amount of energy transferred between two bodies

Therefore, Heat required to evaporate the water to steam is 5342 KJ

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