45.0 g of ice at –10.0 C is mixed with 325 g of water at 37.0 C. Calculate the final temperature of the mixture.

Question

45.0 g of ice at –10.0 C is mixed with 325 g of water at 37.0 C. Calculate the final temperature of the mixture. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/gK, liquid water = 4.18 J/g*K)

The finding the final temperature is throwing me off because that leaves us with so many unknowns. Any help would be great. Thank you

Answer 1

Heat gained by the ice must equal the heat lost by the warm water.

Heat = q = mC∆T or m∆T depending on if there is a phase change or not (see below)

Heat gained by the ice = mC_ice∆T + m∆H_f + mC_water∆T

(45.0g)(2.06 J/g/deg)(10deg) + (45.0g)(333 J/g)(45.0 g)(4.18 J/g/deg)(Tf - 0)

Heat lost by warm water = mC_water∆T

(325 g)(4.18 J/g/deg)(37 - Tf)

Setting these two equations equal to one another we have heat gained by ice = heat lost by warm water...

(45.0g)(2.06 J/g/deg)(10deg) + (45.0g)(333 J/g) + (45.0 g)(4.18 J/g/deg)(Tf - 0) = (325 g)(4.18 J/g/deg)(37 - Tf)

Solving for Tf (final temperature)....

(927) + (14,985) + 188Tf = 50,265 -1386Tf

15912 + 188Tf = 50,265 - 1386Tf

1574Tf = 34,353

Tf = 21.8ºC

(Of course, please check the math)