Part 2: Multi-Step Curves (Day 2) 1. How much energy (in kJ) is required to get a 25.0 g ice cube (water) at -10.0...

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Part 2: Multi-Step Curves (Day 2) 1. How much energy (in kJ) is required to get a 25.0 g ice cube (water) at -10.0 C to compl

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Answer 1

Answer #1

The convert ice at -10⁰c to 0⁰C

q1 = mc $\Delta$ T

= 25*2.09*(0-(-10)

= 25*2.09*10

= 522.5J

The fusion of ice

q2 = m $\Delta$ H fusion

= 25*334 = 8350J

The water convert at 0⁰C to 100⁰C

q3 = mc $\Delta$ T

= 25*4.184*(100-0) = 10460J

The heat of vaporization of water

q4 = m $\Delta$ H

= 25*2257 = 56425J

The steam convert 100⁰C to 110⁰C

q5 = mc $\Delta$ T

= 25*2.09*(110-100) = 522.5J

The total heat energy changes

q = q1 + q2 + q3 + q4 + q5

= 522.5 + 8350 + 10460 + 56425 + 522.5

= 76280J

= 76.28KJ >>>>answer

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Answer 2

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Part 2: Multi-Step Curves (Day 2) 1. How much energy (in kJ) is required to get a 25.0 g ice cube (water) at -10.0...

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Add Answer to: Part 2: Multi-Step Curves (Day 2) 1. How much energy (in kJ) is required to get a 25.0 g ice cube (water) at -10.0...

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