How Much Heat is Needed to Melt 24.0 kg of Ice?

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To determine the heat required to melt 24.0 kg of ice at 0.00°C, the relevant equation is q = M * Lf, where Lf is the latent heat of fusion. Using the provided value of Lf at 334 kJ/kg, the calculation yields q = 24 kg * 334 kJ/kg, resulting in 8.02 * 10^3 kJ. Participants discuss the importance of understanding the symbols Lf and Lv, with one member expressing uncertainty about their meanings. Overall, the correct answer to the problem is 8.02 * 10^3 kJ, confirming the calculation aligns with the expected results. Understanding these concepts is crucial for solving similar heat transfer problems.
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Homework Statement


A 24.0 kg sample of ice is at 0.00°C. How much heat is needed to melt it? (For water Lf = 334 kJ/kg and Lv = 2257 kJ/kg.)

A)5.42*10^4 kJ B)0.00 kJ C)8.02*10^3 kJ D)2.19*10^6 kJ

Homework Equations





The Attempt at a Solution

 
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bigboss, as part of PF forum rules, you need to show some sort of attempt in order for us to help you. (Or else Borek will make you wear that wig! :smile: )

I am sure you know something about this problem...relevant equations.

So, that being said, what do you think about this problem? What kind of concepts have you been talking/reading about in class?

:smile:

P.S. I got you started in your other thread.
 
i believe i have figured this out, q= M*Lf

so 24*334= 8.02*10^3
 
bigboss said:
i believe i have figured this out, q= M*Lf

so 24*334= 8.02*10^3

I am not sure about this. Do you know what you are doing here? Or are you just multiplying numbers until you get one of the offered answers? :smile:

Like Borek asked: what do Lf and Lv stand for? (I know; I just want you to tell me)

EDIT: Been awhile since I have dealt with heat; I think you are correct.
 
Last edited:
Note: I asked, but then I deleted my post to not interfere.

And honestly, I don't know. I can guess, but I have never seen these symbols used. Call it cultural difference :wink:
 

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