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## Homework Statement

An insulated beaker with negligible mass contains liquid water with a mass of 0.325 \rm{kg} and a temperature of 69.9 C.

How much ice at a temperature of -15.0 C must be dropped into the water so that the final temperature of the system will be 39.0 C?

Take the specific heat of liquid water to be 4190 , the specific heat of ice to be 2100 , and the heat of fusion for water to be 334

## Homework Equations

Q = mcT

Q = mL

## The Attempt at a Solution

m = mass of ice

Melt Ice = 334 * m

Heat Ice = 2100 * (0--15) * m

Heat ice water = m*4190 * (39-0)

Cool Water = 0.325 * 4190 (39-69.9) = 42078.075

Solve for m

42078.075 = 334 * m + 2100 * (0--15) * m + m*4190 * (39-0)

m= 0.2155 kg

Found another equation and used it and got the same answer.... not sure where I am going wrong.

Mice = Mwater * Cwater (Tf - Twater)/(Cice * Tice -Lf - Cwater * Tf)

[0.325 * 4190 * (39-69.9)] / (2100 * -15 -334-4190*39) =0.2155kg

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