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

[Finland]

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.

 

[Hootenanny]

Welcome to Physics Forums.

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

 

[Finland]

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) ;-)

 

[Hootenanny]

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).