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?
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