How Much Ice Is Needed to Cool Water from 25°C to 10°C?

AI Thread Summary
The discussion centers on calculating the mass of ice needed to cool 5 liters of water from 25°C to 10°C using latent heat principles. Participants share their calculations, focusing on specific heat values for water and ice, as well as the latent heat of ice. One user initially struggled but eventually arrived at a mass of approximately 0.760 kg, while another confirmed a slightly different result of 0.796 kg. The conversation highlights the importance of understanding heat transfer and energy balance in thermodynamic problems. Overall, the problem-solving process emphasizes collaborative learning in physics.
Finland
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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.
 
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Welcome to Physics Forums.

What are your thoughts on the problem? What have you attempted thus far?
 


Hootenanny said:
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) ;-)
 


Finland said:
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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