- #1

S_fabris

- 20

- 0

A jar of tea is placed in sunlight until it reaches an equilibrium temp of 32.4 deg C. In an attempt to cool the liquid, which has a mass of 185g, 113g of ice at 0degC is added. Assume specific heat capacity of the tea to be that of pure liquid water. At the time at which the temp of the tea is 30.9degC, find the mass of the remaining ice in the jar (in grams)

This is the work i have so far:

Q = heat gained by ice

= mcdeltaT

= 0.113kg x 2090J/kgdegC x (30.9C-0C)

= 7 297.653 J

Q = heat lost by water

= mcdeltaT

= 0.185kg x 4186J/kgdegC x (30.9C-32.4C)

= -1 161.615J

So,

ice from 0degC -> 30.9degC = 7297.653J

water from 32.4degC -> 30.9degC = -1161.615J

In my book it suggests to do Q that is left = Qwater-Qice, this gives me 8459.268J

I really am not sure what do do from here on, any suggestions? I have to find the grams remaining in the jar

Thanks

Sergio