The question is: a 6 pack of cola cans contains m grams of liquid. you want to put ice into a perfectly insulated box to cool it down from Ts to Tf. The ice initially has temp Ti. So (a)how much ice would you need as a minimum. (ignore presence of cans and final state of ice will be water). (b) calculate the mass of ice initially at -5 deg celcius needed to cool 2.1kg of canned cola from 23 deg celcius to 4 deg celcius.
specific heat of water = 4.18J
Lf of water = 333.55Jkg-1
mass(cola) x C (water) (Tf - Ts) = mass(ice) x C(ice) (0 - Ti) + mass(ice) x Lf + mass(ice) x C(water) (Tf - 0)
The Attempt at a Solution
I've worked out that the amount of energy needed to melt 'm'kg of ice = m(ice) x Lf(water) or Q = mL. also that heat lost by cola = heat gained by ice
I realise there are 3 stages involved:
(1) heat needed to raise ice to O deg celcius
(2) heat needed to melt ice
(3) heat needed to raise temp of water (from melted ice) to Tf from O deg celcius.
Then I get lost. How do I find the mass of the ice to begin with? Do I rearrange the equation? If so, I have no idea how to start rearranging a long complex equation like this.
Thank you for any guidance on this.