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ngtbma
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Joined: 30 Apr 2005
Posts: 17
Location: CA
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 Posted: Sun, 7 Jan 2007 04:35:37 UTC Post subject: heat required for phase change |
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Hi I am having trouble calculating this problem:
How much heat must be added to bring 20g of ice at 0 C to water vapor at 100 C?
Heat of fusion of ice: 80 cal/g
Heat of vaporization of water is: 540 cal/g
The method that I used is
q=m * delta C = (20g)(80cal/g)=1600cal for heat of fusion
q=m*delta C* delta T = (20g)(540cal/g)(100C)= 1080000 cal for heat of vaporization...so when i added the two I got 1081600 cal. However it is wrong, but I don't know how to get 14400 cal, which is the answer in the book. Can you show me the steps to getting that answer. Thank you. |
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Unco
Member of the 'S.O.S. Math' Hall of Fame
Joined: 07 Jan 2005
Posts: 1142
Location: Christchurch, New Zealand
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| Posted: Sun, 7 Jan 2007 04:47:04 UTC Post subject: Re: heat required for phase change |
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| ngtbma wrote: | Hi I am having trouble calculating this problem:
How much heat must be added to bring 20g of ice at 0 C to water vapor at 100 C?
Heat of fusion of ice: 80 cal/g
Heat of vaporization of water is: 540 cal/g
The method that I used is
q=m * delta C = (20g)(80cal/g)=1600cal for heat of fusion
q=m*delta C* delta T = (20g)(540cal/g)(100C)= 1080000 cal
Always check your units make sense.
for heat of vaporization...so when i added the two I got 1081600 cal. However it is wrong, but I don't know how to get 14400 cal, which is the answer in the book. Can you show me the steps to getting that answer. Thank you. |
To change from solid to liquid, 1600cal is needed as you said. Similarly, 10800cal is needed to change from liquid to gas. In between, the temperature is raised by 100C which, for 20g of liquid water (specific heat capacity 1cal/g/C), needs 2000cal. Add them up to get 14400cal. |
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ngtbma
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Joined: 30 Apr 2005
Posts: 17
Location: CA
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| Posted: Sun, 7 Jan 2007 05:04:51 UTC Post subject: |
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| ahhh....I get it now...thank you for replying my problem so quick. |
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