# Change in Entropy

1. Feb 6, 2008

### hils0005

A 10g ice cube at -10C is placed in a lake whose temp is 15C, calculate the change in entropy of the cube-lake system as the ice comes to thermal equilibrium with the lake.
specific heat of ice is 2220 J/KgK

2. Relevant equations
Q(released)=Q(absorbed)
U=Q-W
S=Q/T
L(f)water=333kJ/kg

3. The attempt at a solution
Not realy sure how to start??
Do you need to calculate change in the ice from -10C to 0c, then from 0C ice to 0C water, 0C water to 15C?
1. Q(ice)=c(ice)M(ice)(Tf-Ti)
Q=(2220 J/kgK)(0.01kg)(273K-263K)=222J

2.Q(ice to water)=Lm=333kJ/kg(.01kg)=3.33kJ=3330J ?

3.Q(water)=c(w)m(w)(Tf-Ti)
Q=(4190J/kgK)(.01kg)(288K-273K)=628.5J

Q=222+3330+628.5=4180.5J

I don't know what to do now or even know if I'm on the right path

2. Feb 6, 2008

3. Feb 7, 2008

### hils0005

this is a different question

4. Feb 7, 2008

### Andrew Mason

Yes, you are right. Sorry. You had double posted the earlier one though.

To calculate the change in entropy you have to work out the change in entropy of the ice cube water and the change in entropy of the lake, and add the two.

The change in entropy of the ice cube/water is the integral:

$$\Delta S = S_b - S_a = \int_a^b dQ/T = cm \int_a^b dt/T = cm\ln\frac{T_b}{T_a}$$

Divide the process into three separate stages: -10 to 0; melting (at 0C); 0 to 10 C because c is different for each of those stages.

To find the change in entropy of the lake (is it positive or negative?) find the total heat flow (from the lake to the ice cube/water) and divide by the lake temperature.