Find the changes in entropy for the water and for the ice

[Soaring Crane]

In a well-insulated calorimeter, 1.0 kg of water at 20 C is mixed with 1.0 g (.001 kg) of ice at 0 C. What is the net change in entropy of the system by the moment the ice completely melts? The heat of fusion of ice is 3.34 x 10^5 J/kg.


I must find the changes in entropy for the water and for the ice, then find the net change.

Net change = /_\S1 + /_\S2

This is what I did so far:
S_ice = (mL)/T, where Q = mL

S_ice = [(3.34 x 10^5 J/kg)*.001 kg]/[273 K] = 1.223 J/K ?

How do I find the change in entropy for H2O??

Thanks.

 

[Andrew Mason]

In a well-insulated calorimeter, 1.0 kg of water at 20 C is mixed with 1.0 g (.001 kg) of ice at 0 C. What is the net change in entropy of the system by the moment the ice completely melts? The heat of fusion of ice is 3.34 x 10^5 J/kg.

This is what I did so far:
S_ice = (mL)/T, where Q = mL

S_ice = [(3.34 x 10^5 J/kg)*.001 kg]/[273 K] = 1.223 J/K ?

How do I find the change in entropy for H2O??

You have found the $\Delta S$ for the ice. The $\Delta Q$ for the water is just the loss of heat to the ice. It takes place at 293K (we can see that there is not a significant temperature reduction - about .1 degree). So:

$$\Delta S_{H_2O} = -334/293 = -1.14 J/K$$

AM

 

[wais]

To find the change in entropy for water, we can use the equation S = Q/T, where Q is the heat absorbed or released and T is the temperature in Kelvin. In this case, the water is being heated from 20 C (293 K) to 0 C (273 K), so the change in entropy for water would be:

S_water = Q/T = [(1 kg)(4186 J/kg*K)(273 K - 293 K)]/293 K = -13.6 J/K

Since the ice is at 0 C and the water is at 20 C, we can assume that the heat lost by the water is equal to the heat gained by the ice. Therefore, the net change in entropy for the system would be:

Net change = S_water + S_ice = -13.6 J/K + 1.223 J/K = -12.377 J/K

So the net change in entropy for the system is -12.377 J/K.