Hw 7.37

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404324273
Posts: 4
Joined: Tue Nov 25, 2014 3:00 am

Hw 7.37

Postby 404324273 » Tue Jan 13, 2015 10:29 pm

How much heat is needed to convert 80.0g of ice at 0 degrees into liquid water at 20 degrees?

So I tried using q=mCs(delta T).. i know i'm suppose to use the specific heat capacity of water and the standard enthalpy of fusion..but i'm not sure how to combine the two to solve for q?

Sanmeet Atwal 1D
Posts: 17
Joined: Fri Sep 26, 2014 2:02 pm

Re: Hw 7.37

Postby Sanmeet Atwal 1D » Tue Jan 13, 2015 11:33 pm

There are two parts to this problem since you need to melt the 80.0 grams of ice at 0.00 degrees C and then heat the liquid water to 20 degrees C. So you need to find the heat needed in order to melt the ice and add it to the heat needed to heat the water to 20 degrees C.
1.) In order to find ΔH of melting the ice you need to use ΔH(fusion) of water which is found on the charts: 6.01 Kj/mol
You know that ΔH = molar/mass * ΔH(fusion) = (80g/18.02g/mol)*(6.01kJ/mol)= 26.7 kJ of heat needed to melt the ice.
2.) In order to find the heat needed to raise the temperature of the liquid you use: q=mCΔT= (80.0g)(4.187J/C/g/)(20C-0C)=6.69kJ of heat needed to raise the temperature.

Finally, you add the two values of heat together in order to find the total heat needed for the entire process.
= 26.7 kJ + 6.69 kJ = 33.4 kJ of total heat needed
Hope this helps! (:

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Hw 7.37

Postby Niharika Reddy 1D » Tue Jan 13, 2015 11:33 pm

You can split the problem into 2 parts, finding the change in enthalpy associated with 1. melting the ice and adding it to the change in enthalpy of 2. raising the temperature of the liquid water to its final temperature: 20°C.

For melting the ice, multiply the standard enthalpy of fusion (6.01 kJ/mol) by the number of moles of ice present. The "mol" unit will cancel, leaving you with the heat required to melt the ice.

To raise the temperature of liquid water, use q=mCΔT which will give the heat needed in kJ.

Now add these 2 values together to get the total heat required.


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