Energy lost in conversion of steam to ice

[ttk3]

Homework Statement

How many J of energy must be removed when 145.0 g of steam, at a temperature of 188.0°C, is cooled and frozen into 145.0 g of ice at 0°C? Take the specific heat of steam to be 2.1 kJ/(kg·K).

Homework Equations

Q=m*L

The Attempt at a Solution

Energy = Mgas:liquid*Lvaporization + Mliquid:solid * Lfusion

.145*22.6E5 + .145*33.5E4 = 3.76E5 J

 

[mgb_phys]

You also need to account for the energy to cool the steam from 188C to and then cool the water from 100C to 0C.
Use Q = m*c*T

 

[ttk3]

hm... it's still not working.

I added .145(2.1)(88) + .145*4186*100 = 60723.769 J

The total I found was 4.37E5 J.

 

[mgb_phys]

cool steam = 0.145 * (188-100) * 2100
condense steam = 0.145 * 2272E3
cool water = 0.145 * (100-0) * 4181
freeze water = 0.145 * 334E3

Just a small typo, the heat capcity of steam is 2.1KJ/kg = 2100 J/kg