Calculate Heat Removed from 130 g Steam to Ice

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SUMMARY

The discussion focuses on calculating the total heat removed when 130 g of steam at 145°C is cooled and frozen into ice at 0°C. The specific heat of steam is given as 2.01 kJ/kg·K, and the specific heat of water is 4.186 J/g·°C. The user initially calculated the heat removed during cooling and phase change but encountered an error likely due to neglecting the latent heat of condensation. The correct approach requires incorporating the latent heat of vaporization for steam to water before further cooling.

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Homework Statement



How much heat must be removed when 130 g of steam at 145°C is cooled and frozen into 130 g of ice at 0°C. (Take the specific heat of steam to be 2.01 kJ/kg·K.)

Answer is in kcal.

Homework Equations



sht.gif


The Attempt at a Solution



I know steam turns back into water at 100°C

Q1=C1m(deltaT)
Q1=(2010 J//kg·K)(.13kg)(418.15K-373.15K)

In the second part, instead of using the specific heat of steam, I used the specific heat of water= 4.186 joule/gram °C

Q2=C2m(deltaT)
Q2=(4.186 J/g·°C)(130g)(100°C)

I then added Q1 to Q2 and came up with an answer in Joules.
1 joule = 0.000239005736 kilocalories

But my answer is wrong. Anyone see what I missed? Probably a conversion issue.
 
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Condensation (latent heat).
 

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