Determine the total heat (kJ) required to condense 2.5 x 105 kg of stead at 100.0oC into water and then cool the water from 100.0oC to 0.0oC.

Question

(C_steam = 2.03 J/(g*^oC), C_water = 4.18 J/(g*^oC), C_ice = 2.06 J/(g*^oC), ∆H_vap = 2260 J/g, and ∆H_fus = 334 J/g) 

Answer 1

I'm guessing that stead is supposed to be steam. I'm also guessing that 2.5x105 kg might be 2.5x105 g, but I'll go ahead and answer it as written which represents an enormous mass of water and hence an enormous amount of heat energy.

We will use the general formula of q = mC∆T to answer this question

q = heat

m = mass =

C = specific heat

∆T = change in temperature.

Step 1: convert 2.5x10⁵ kg of steam @100º to 2.5x10⁵ kg of liquid H2O @ 100º (phase change). For a phase change there is no change in temperature and we use ∆H instead of C. So, we have...

q = (m)(∆H_vap) = (2.5x10⁵ kg)(2260 kJ/kg) = 5.65x10⁸ kJ

Step 2: convert 2.5x10⁵ kg of liquid H2O at 100º to liquid H2O at 0º.

q = (2.5x10⁵ kg)(4.18 kJ/kg)(100 deg) = 1.05x108 kJ

Add them up to get a total of 5.93x10⁸ kJ total heat needed to bring the steam to liquid water at 0º