Total entropy change for ice melting in a room

[plpg]

Homework Statement

Okay, so I am having difficulties with understanding the concepts around entropy, take this question:
What is the total entropy change for 7 kg of ice melting from -5 C° to 5 C° in room at 5 C°.

Homework Equations

dS=dQ/T
Q =mΔH
m*c*ln(tfinal/tinitial)
c_ice=2
c_water=4
c_air=1

The Attempt at a Solution

Okay, so this is my reasoning so far:
as the ice heats up to zero centigrades, the total change in entropy is given by integrating dS=dQ/T wrt T, and we end up at : m*c*ln(tfinal/tinitial), so 7*2*ln(273.15/(273.15-5)) j/k.

As the ice changes its state to water, the change in entropy is given directly by dS=dQ/T, where dQ is =mdH=7*330=2310 j/k.

When the water heats up from 0 centigrades to 5 centigrades its change in entropy is again given by, 7*4*ln(273.15+5/(273.15)) j/k,
and adding these we get the total entropy change for the melting ice.

Now to the part that I am having big difficulties understanding: the change in entropy of the room. I've read that it is given by m*c_air*((tfinal-tinitial)/tfinal), so in this case this would be negative 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

So adding all these would give me the answer? 7*2*ln(273.15/(273.15-5)) j/k + 2310 j/k + 7*4*ln(273.15+5/(273.15)) j/k - 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

Am i even close to getting this question right?

 

[Andrew Mason]

Okay, so this is my reasoning so far:
as the ice heats up to zero centigrades, the total change in entropy is given by integrating dS=dQ/T wrt T, and we end up at : m*c*ln(tfinal/tinitial), so 7*2*ln(273.15/(273.15-5)) j/k.

This is correct except for the units. What are the units for the specific heat of ice?

As the ice changes its state to water, the change in entropy is given directly by dS=dQ/T, where dQ is =mdH=7*330=2310 j/k.

Again, correct except for the units.

When the water heats up from 0 centigrades to 5 centigrades its change in entropy is again given by, 7*4*ln(273.15+5/(273.15)) j/k,
and adding these we get the total entropy change for the melting ice.

Units.

Now to the part that I am having big difficulties understanding: the change in entropy of the room. I've read that it is given by m*c_air*((tfinal-tinitial)/tfinal), so in this case this would be negative 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

So adding all these would give me the answer? 7*2*ln(273.15/(273.15-5)) j/k + 2310 j/k + 7*4*ln(273.15+5/(273.15)) j/k - 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

Am i even close to getting this question right?

You have to assume that the temperature of the room air does not change. The room is large enough that the heat flow from the air changes the temperature of the air by a negligible amount. So the entropy change is just$\Delta S_{room} = \Delta Q_{room}/T_{room}$ where $\Delta Q_{room} = - \Delta Q_{ice}$.
AM

 

[plpg]

1234

 

[Andrew Mason]

Thank you for your reply. I make the units out to be Joule per kelvin, is that something i should look more into or was it the fact that I wrongly used small j and k?

If you are using Kg and a specific heat of 2 KJ/Kg for ice and 4 KJ/Kg for water (which I assume is the case since you are using 7 and not 7,000 as the mass), the units for entropy will be in KJ/K.

"You have to assume that the temperature of the room air does not change. The room is large enough that the heat flow from the air changes the temperature of the air by a negligible amount." Aha okay, So the total change in entropy ought to be more in the lines of :
S_total=S_Ice-(ΔQ_ice/T_room)

Correct so far.

= (7*2*ln(273.15/(273.15-5))+ 2310+ 7*4*ln(273.15+5/(273.15)))J/K - (ΔQ_ice/(273.15+5)) ?
Where ΔQ_ice=mc_iceΔT+mΔH+mc_waterΔT i.e the total energy required to heat 7 kg of ice from -5c° to 5c° ?

If you are using mass in Kg, the specific heat is in units of KJ/Kg. For example, the total heat flow required to melt 7kg of ice is 7*330 KJ = 2310 KJ = 2,310,000 J.

AM

 

[berkeman]

Please remove, I see that there are many similar threads already on this.

1234

Cheating by deleting your posts after you have received help is strictly against the PF rules. Check your PMs, and do not do this again here.