- #1

markdflip

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## 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

## Homework Equations

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.