Need help for my Final.?
2017-06-07 1:34 pm
A container with 150.0 g of water at 25.8°C is placed in a refrigerator. Eventually, the water becomes ice at –5.0 °C. What is the quantity of heat removed from the water.
回答 (1)
2017-06-07 2:51 pm
Refer to: http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
Heat capacity of water = 4.187 J / (g °C)
Heat capacity of ice = 2.108 J / (g °C)
Latent heat of fusion = 334 J / g
Energy removed for (150.0 g of water at 25.8°C) to (150 g of water at 0°C)
= 150.0 × 4.187 × (25.8 - 0) J
Energy removed for (150.0 g of water at 0°C) to (150.0 g of ice at 0°C)
= 150.0 × 334 J
Energy removed for (150.0 g of ice at 0°C) to (150.0 g of ice at -5.0°C)
= 150.0 × 2.108 × [0 - (-5.0)] J
Total energy removed for ((150.0 g of water at 25.8°C) to (150.0 g of ice at -5.0°C)
= 150.0 × 4.187 × (25.8 - 0) + 150.0 × 334 + 150.0 × 2.108 × [0 - (-5.0)] J
= 67900 J
Heat capacity of water = 4.187 J / (g °C)
Heat capacity of ice = 2.108 J / (g °C)
Latent heat of fusion = 334 J / g
Energy removed for (150.0 g of water at 25.8°C) to (150 g of water at 0°C)
= 150.0 × 4.187 × (25.8 - 0) J
Energy removed for (150.0 g of water at 0°C) to (150.0 g of ice at 0°C)
= 150.0 × 334 J
Energy removed for (150.0 g of ice at 0°C) to (150.0 g of ice at -5.0°C)
= 150.0 × 2.108 × [0 - (-5.0)] J
Total energy removed for ((150.0 g of water at 25.8°C) to (150.0 g of ice at -5.0°C)
= 150.0 × 4.187 × (25.8 - 0) + 150.0 × 334 + 150.0 × 2.108 × [0 - (-5.0)] J
= 67900 J
收錄日期: 2021-04-24 00:28:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170607053452AAc6jpC