Ice cooling tea from 30.1°C to 27.3°C

[Aikenfan]

Please Help Ice and a Jar of Tea

Homework Statement

Given; The specific heat of water is 4186 J/kg deg Celsius.
A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30.1 deg Celsius. In an attempt to cool the liquid, which has a mass of 187g, 97.8g of ice at 0 deg Celsius is added. Assume the specific heat capacity of the tea to be that of pure liquid water. At the time at which the temperature of the tea is 27.3 deg Celsius, find the mass of the remaining ice in the jar. Answer in units of g

Homework Equations

Q = mc (change in T)

The Attempt at a Solution

Q = mc (change in T) = -mc (change in T)
(.187)(4186)(30.1-27.3) = -m (4186)(30.1)
= 17.4g

but i get the answer wrong when i put it into the computer...if anyone can help me with what i am doing wrong, it would be greatly appreciated!

 

[denverdoc]

There is a phase change which you have not accounted for. Even tho there is no change in temperature, "melting ice" or boiling water takes additional energy. the algebra is a little tricky therefore, but doable with careful use of variables, and the assumption I believe you need to make that the remaining ice is still at zero degrees. So one end of the eqn looks somthing like, assuming x to be the fraction melted and warmed:

(97.8-x)(deltaT {diff between 0 degrees and final}+mystery factor alluded to above)= deltaT (which is the diff beteen its initial temp and final temp) times mass of tea.

see if that helps.