posted by Jini on .
How much heat (in BTU) would be required to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm? Assume heat capacity is constant at 1 BTU/lb-°F and the latent heat of vaporization is 1,000 BTU/lb. Also, what is the temperature of the saturated steam at 1 atm?
Is that 25C or 25F? I will assume it is F but if not you can change from F to C and the reverse.
q1 = heat to move water from 25 F to 212F.
q1 = 100 lb x 1 BTU/lb*F x (Tfinal-Tinitial) where Tf-Ti = 212-25
q2 = heat to vaporize water at 212 F to steam at 212 F.
q2 = 100 lb x 1000 BTU/lb.
Total Q = q1 + q2.
The second question answer is in my q2 statement.