Posted by Vladimir on Friday, July 20, 2012 at 11:11am.
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.
The specific heat of ice and water is given as 2.06 kJ/(kg K) and 4.19 kJ/(kg K) respectively.
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.
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.
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
Thank you both.
Tf= 29.30 degree celsius
Tf= 29.30 degree celsius
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