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Heat Removed |
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| May11-10, 10:35 AM | #1 |
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Heat Removed
1. The problem statement, all variables and given/known data
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. 2. Relevant equations ![]() 3. 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. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
| May11-10, 12:40 PM | #2 |
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Condensation (latent heat).
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