Posted by **Vladimir** on Friday, July 20, 2012 at 11:11am.

A 6-g ice cube at −8°C is dropped into 70 g of water at 31°C.

(a) After enough time has passed to allow the ice cube and water to come into equilibrium, what is the temperature of the water?

(b) If a second ice cube is added, what will the temperature be?

- Physics -
**Vladimir**, Friday, July 20, 2012 at 11:13am
The answers are 21.9 degree c and 14.2 degree c. I've tired figuring out a way to use the dE = Q - W equation to solve this problem, but it hasn't been working.

- Physics -
**Vladimir**, Friday, July 20, 2012 at 11:14am
The specific heat of ice and water is given as 2.06 kJ/(kg K) and 4.19 kJ/(kg K) respectively.

- Physics -
**ajayb**, Friday, July 20, 2012 at 12:13pm
You might have forgotten to include the latent heat - heat required to convert the state from ice to water w/o change in temp. Consider this factor and you could get the correct answer.

- Physics -
**Vladimir**, Friday, July 20, 2012 at 1:13pm
I'm not too sure how to add the latent heat part in there, but I'll reread that section in my book and see if I can figure something out.

- Physics -
**Elena**, Friday, July 20, 2012 at 3:39pm
m1=0.006 kg, T1= - 8ºC, m2=0.07 kg, T2=31 ºC

m1•c(ice) •(0-(-8)+ m1•L+m1•c(water)(T-0) = m2•c(water)(31-T)

0.006•2060•8 + 0.006•335000+0.006•4190•T)=0.07•4190(31-T)

98.88+2010+25.14T=9092.3 – 293.3T

T(25.14+293.3)= 9092.3-98.88-2010

318.44 T=6983.42

T=21.93ºC.

New equation (m3=m1+m2=0.07+0.006=0.076 kg)

m1•c(ice) •(0-(-8)+ m1•L+m1•c(water)(T1-0) = m3•c(water)(21.93-T1)

Solve for T1

- Physics -
**Vladimir**, Friday, July 20, 2012 at 7:07pm
Thank you both.

- Physics -
**Sherileen Eva**, Sunday, November 18, 2012 at 5:55am
Tf= 29.30 degree celsius

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