# Latent heat calculation

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
Staff Emeritus
Gold Member

Welcome to Physics Forums.

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

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
Staff Emeritus
Gold Member

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).