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### Question regarding 8.39 homework question

Posted: **Wed Jan 24, 2018 5:27 pm**

by **Michelle Chernyak 1J**

The question (8.39) is "how much heat is needed to convert 80.0 g of ice at 0.0C into liquid water at 20.0C?"

I understand how to do the problem but I don't understand why we need to employ the specific heat capacity of liquid water rather than ice. Can someone please explain that to me?

### Re: Question regarding 8.39 homework question

Posted: **Wed Jan 24, 2018 5:40 pm**

by **Tim Foster 2A**

Hey Michelle,

You need to use the specific heat capacity of liquid water because once the ice melts, all heat being transferred will be subject to the properties and heat capacity of liquid water. This problem can be essentially be split into two parts; the first of which you use the heat capacity of solid ice to figure out how much energy is needed to make the ice melt, and the second of which is figuring out how much heat needs to be transferred to your now-liquid water to raise its temperature from "0" degrees to 20 degrees, for which you use the specific heat capacity of liquid water.

### Re: Question regarding 8.39 homework question

Posted: **Fri Jan 26, 2018 12:12 am**

by **Joanne Guan 1B**

When ice is melting, the temperature doesn't change until it is fully melted. That means you can only calculate the change from 0C to 20C until the ice is fully melted and becomes water. Therefore, you have to use the specific heat capacity of water.

Now, if the ice were at -10C, then you would calculate the change in temperature from -10C to 0C using the specific heat capacity of ice, and then continue on with the above.

### Re: Question regarding 8.39 homework question

Posted: **Sat Jan 27, 2018 6:02 pm**

by **Diane Bui 2J**

Adding on above, I think it's useful to draw a diagram of the heating curve to understand what happens first. In this case, the ice melts to liquid water first and then the temperature is raised. Therefore, the specific heat capacity of liquid water instead of solid water.

### Re: Question regarding 8.39 homework question

Posted: **Sat Jan 27, 2018 11:56 pm**

by **Oscar Valdovinos 1I**

In short, this problem requires two steps, the first step is calculating how much heat is needed to melt the ice at 0c, basically asking you to calculate the phase change with 80grams of ice (DeltaH=n(6.01kj*mol). The second part is calculating a change in temperature with the 80g of liquid water which is why we have to q=mcat for liquid water because the reaction is taking place in the temps where the phase of is liquid. It is definitely helpful to draw enthalpy graph associated with water to help visualize the problem.

### Re: Question regarding 8.39 homework question

Posted: **Sun Jan 28, 2018 10:19 pm**

by **Kayla Tchorz-Dis 1F**

The way I like to think of it is that you want to use the specific heat of the phase that it is changing into, as long as you're temperature threshold does not cross between two phases. This would do this if the problem would have started with -1 degree celsius. Hope that helps!