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First convert 1 lb of water to lb-moles which is 0.055 lb-moles (you'll need this later). This problem can be broken into 3 steps:

(1) You need to detemine how much heat is needed to raise room temperature water (68oF) to 212oF. This can be used using the heat capacity of water which at room temperature is 1 Btu/lboF. So the amount of heat needed for this is:

Q1 = m*Cp*ΔT

= (1 lb)*(1 Btu/lboF)*(212 - 68oF)

= 144 Btu

(2) Next you need to account for the phase change. The water changes to steam at 212oF. You use the heat of vaporization which you can look up in any Chemistry or Chemical Engineering Handbook. The Hvap that I found is 17493.5 Btu/lb-mole.

Q2 = n(lb-moles)*Hvap

= (0.055 lb-moles)*(17493.5 Btu/lb-mole)

= 972.64 Btu

(3) Next you need to find out how much heat is needed to raise the temperature of the steam from 212 to 213oF. You can look up the heat capacity of steam at 212oF to be 0.485 Btu/lboF.

Q3 = m*Cp*ΔT

= (1 lb)*(0.485 Btu/lboF)*(213-212oF)

= 0.485 Btu

To find the total heat needed add Q1+Q2+Q3 (144+972.64+0.485) =

1117.12 Btu

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