Energy lost in conversion of steam to ice

AI Thread Summary
To determine the energy removed when cooling and freezing 145.0 g of steam at 188.0°C to ice at 0°C, the specific heat of steam is 2.1 kJ/(kg·K). The calculations involve cooling the steam to 100°C, condensing it to water, cooling the water to 0°C, and then freezing it. The total energy calculated includes contributions from each phase change and temperature drop, resulting in approximately 437,000 J. A correction was noted regarding the specific heat of steam, clarifying it as 2100 J/kg. Accurate calculations are crucial for understanding energy conversion in this process.
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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
 
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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
 
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.
 
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
 
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