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**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!

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