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Question 8.39 (Sixth Edition)

Posted: Mon Feb 11, 2019 12:33 am
by Steve Magana 2I
Question: How much heat is needed to convert 80.0 g of ice at 0.0 degrees Celsius into liquid water at 20.0 degrees Celsius?

I'm having trouble understanding this question and what to do. Thank you!

Re: Question 8.39 (Sixth Edition)

Posted: Mon Feb 11, 2019 10:23 am
by Tam To 1B
For this problem, you have to calculate the enthalpy of fusion (ice melting) and the heat of the water to determine how much needs to be supplied in total.
The enthalpy of fusion, delta H, can be calculated using the value from table 8.3. Convert 80.0 g H20 to moles and multiply it with the enthalpy of fusion 6.01 kJ/mol to get 26.7 kJ.
q(h2o) = mCdeltaT = (80.0 g)(4.184 J/gC)(20-0 C) = 6694 J = 6.69 kJ.
Together, 26.7 + 6.69 = 33.4 kJ.

Re: Question 8.39 (Sixth Edition)

Posted: Mon Feb 11, 2019 2:44 pm
by Amy Dinh 1A
You have to break this problem up into two steps:

1. Find m * Heat of fusion
2. Find m * C * delta T

When you find those two values, you add them up, and that is the total heat needed.

Re: Question 8.39 (Sixth Edition)

Posted: Mon Feb 11, 2019 11:06 pm
by Jennifer Su 2L
Amy Dinh 1A wrote:You have to break this problem up into two steps:

1. Find m * Heat of fusion
2. Find m * C * delta T

When you find those two values, you add them up, and that is the total heat needed.


Is it n*Heat of fusion or m* Heat of fusion? For the first part we use moles and the second part we use mass right?

Re: Question 8.39 (Sixth Edition)

Posted: Fri Feb 15, 2019 2:42 pm
by Vanadium Wang 4H
Jennifer Su 2L wrote:
Amy Dinh 1A wrote:You have to break this problem up into two steps:

1. Find m * Heat of fusion
2. Find m * C * delta T

When you find those two values, you add them up, and that is the total heat needed.


Is it n*Heat of fusion or m* Heat of fusion? For the first part we use moles and the second part we use mass right?

It would be n*Heat of fusion, you're right. Heat of fusion is given in kJ/mol so in order to obtain kJ you would need to multiply by mol. So for the first part we, indeed, use moles and mass for the second part.