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

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Homework Help Overview

The problem involves calculating the mass of ice needed to cool a specific volume of water from 25°C to 10°C, considering the specific heat capacities of water and ice, as well as the latent heat of fusion for ice. The scenario assumes no heat exchange with the environment.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster seeks hints for approaching the problem, indicating a focus on the calculation aspect. Some participants share their calculations and results, while also questioning the accuracy of their answers.

Discussion Status

Participants have engaged in sharing their thought processes and calculations. There is a recognition of differing results, but no explicit consensus on the correct mass of ice has been reached. The discussion appears to be productive, with participants offering insights into their reasoning.

Contextual Notes

The original poster mentions a translation from Finnish, which may introduce some ambiguity in the problem statement. Additionally, the problem is framed within the context of a homework assignment, suggesting constraints on the approach to be taken.

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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