Moment the ice completely melts?

[~angel~]

Please help.

In a well-insulated calorimeter, 1.0 kg of water at 20C is mixed with 1.0g of ice at 0C. 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*10^5 J/kg.

I'm not sure how to do this problem. I thought that since it is 1kg of water, change in entropy for that is -4190/293K. I also though that for the ice, it would be (0.001*3.34*10^5)/273 as well as 4190/273 when it melts.

Thank you.

 

[Andrew Mason]

Please help.

In a well-insulated calorimeter, 1.0 kg of water at 20C is mixed with 1.0g of ice at 0C. 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*10^5 J/kg.

I'm not sure how to do this problem. I thought that since it is 1kg of water, change in entropy for that is -4190/293K. I also though that for the ice, it would be (0.001*3.34*10^5)/273 as well as 4190/273 when it melts.

The change in entropy of the water is:

$$\Delta S_{water} = \Delta Q/T_{water}$$

where $\Delta Q$ is the heat loss, which is the same as the heat gained by the ice.

AM

 

[~angel~]

Would Q for water just be -4190, or is there something else involved?
I'm also not sure about the ice.

 

[janiexo]

You've already calculated the change of Q of the ice (0.001*3.34*10^5). So the change in entropy of the ice is just (0.001*3.34*10^5)/273, which you've done.
Regarding the change in entropy of the water, the change in Q is just the negative of the Q gained by the ice i.e. it would be -(0.001*3.34*10^5). So the change in entropy of the water is just -(0.001*3.34*10^5)/293. So now you can calculate the net change in entropy by just adding them up :)
Hope you understood that, I'm horrible at explaining things

 

[~angel~]

Yeah, thank you =)