Question 8.39 (Sixth Edition)

[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!

[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.

[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.

[Jennifer Su 2L]

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?

[Vanadium Wang 4H]

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.