Question

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?

Answer 1

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.

Answer 2

The specific heat of ice and water is given as 2.06 kJ/(kg K) and 4.19 kJ/(kg K) respectively.

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

Thank you both.

Answer 7

Tf= 29.30 degree celsius