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Latent heat calculation

  • Thread starter Finland
  • Start date
  • #1
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Latent heat calculation!!

Ok... first timer. I am not a science student, done some MBA but wanting some IT course and for entrance exam, I am preparing. I have problem with this one problem. Just provide me a hint. I do calculation part.

'Juice and water is produced from 5.0 liters of water, a temperature of 25 ° C. Ice is added of which temperature is -18 oC. How large ice cube mass should be in order to drink, the final mixing temperature is 10 ° C? Suppose
to drink
and the environment, there is no heat exchange. Juice and water
specific heat is 4.19 kJ / (kg ° C), ice specific heat 2.2 kJ / (kg ° C) and ice
latent heat is 333 kJ / kg. Juice and water density is 1.0 kg/dm3.'

(i just translated this question from finnish to english, so grammar mistake is there!!)

Please do provide me hints as to how I proceed with this question.
 

Answers and Replies

  • #2
Hootenanny
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Welcome to Physics Forums.

What are your thoughts on the problem? What have you attempted thus far?
 
  • #3
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Welcome to Physics Forums.

What are your thoughts on the problem? What have you attempted thus far?
ha ha 2 hours i attempted this morning... didn't care about breakfast... now i just got it right!!

Anyway.. kiitos (thanx)

goes this way
(1kg/dm3 x 5 l) x 4.19 kJ/kgC x 15 = 2.2 kJ/kgC x m x 28 + 333 kJ/kg x m

answer is, m = 0.760 kg (but i got somewhere 0.796) ;-)
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6


ha ha 2 hours i attempted this morning... didn't care about breakfast... now i just got it right!!

Anyway.. kiitos (thanx)

goes this way
(1kg/dm3 x 5 l) x 4.19 kJ/kgC x 15 = 2.2 kJ/kgC x m x 28 + 333 kJ/kg x m

answer is, m = 0.760 kg (but i got somewhere 0.796) ;-)
My answer is the same as yours (0.796kg).
 

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