# Energy added to Ice- determine final temperature

markdflip

## Homework Statement

A 10 kg block of ice has a temperature of -8°C. The pressure is one atmosphere. The block absorbs 4.17 x10^6 J of heat. What is the final temperature of the liquid water?

Q=4.17x10^6J
T_0=-8
Mass=10kg

4184= specific heat of water
2000=specific heat of ice
333.5=latent heat of fusion

Q=mc(deltaT)
Q=mL

## The Attempt at a Solution

Q_1=mc(delta)T
Q_1=(10kg)(2000)(8°C)
Q_1=160,000J of energy was needed to raise the temperature to 0 degrees Celsius

Q_2=mL
Q_2=(10kg)(333.5)
Q_2=3335J of energy needed to change ice to water.

160,000J+3335J=163,335J of energy used.
4,170,000J-163,335J= 4,006,665J of energy left

4,006,665J=mc(delta)T
4,006,665J=(10kg)(4184)(delta)T
4,006,665/(10*4184)=deltaT
deltaT=95.761°C
95.761 degrees Celsius should be the final temperature according to my calculations.
I'm fairly sure the process is right, my money is there is something wrong with a coefficient, but I can't figure it out.

Homework Helper
check your units for the heat of fusion

markdflip
oh, that was in J/g...so that would be 333,550 J/Kg
That makes more sense, 674,500J would be left for the third step opposed to 4,006,665J

674,500J/(10*4184)=deltaT
DeltaT=16.12 degrees Celsius.
Sound realistic?