- #1
bigboss
- 28
- 0
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
In summary, the problem asks to find the amount of heat needed to melt a 24.0 kg sample of ice at 0.00°C, given that for water Lf = 334 kJ/kg and Lv = 2257 kJ/kg. The correct answer is C) 8.02*10^3 kJ, which is calculated by multiplying the mass of the ice by the latent heat of fusion (Lf = 334 kJ/kg).
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
- #2
Saladsamurai
- 3,020
- 7
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! 
)
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?
P.S. I got you started in your other thread.
)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?
P.S. I got you started in your other thread.
- #3
bigboss
- 28
- 0
i believe i have figured this out, q= M*Lf
so 24*334= 8.02*10^3
so 24*334= 8.02*10^3
- #4
Saladsamurai
- 3,020
- 7
bigboss said:
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?
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:
- #5
Borek
Mentor
- 28,961
- 4,266
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
And honestly, I don't know. I can guess, but I have never seen these symbols used. Call it cultural difference
Related to How Much Heat is Needed to Melt 24.0 kg of Ice?
1. How much heat is needed to melt 24.0 kg of ice?
The amount of heat needed to melt 24.0 kg of ice can be calculated using the formula Q = m*L, where Q is the heat required, m is the mass of ice, and L is the specific latent heat of fusion for ice. The specific latent heat of fusion for ice is 334 kJ/kg. Therefore, the heat needed to melt 24.0 kg of ice is 334 kJ/kg * 24.0 kg = 8016 kJ.
2. How long will it take to melt 24.0 kg of ice?
The time it takes to melt 24.0 kg of ice will depend on the rate at which heat is applied. If we assume a constant rate of heat transfer, we can use the equation Q = m*L*t, where t is the time in seconds. Using the same values as before, we can calculate the time to be t = Q / (m*L) = 8016 kJ / (24.0 kg * 334 kJ/kg) = 0.95 hours or approximately 57 minutes.
3. What factors can affect the amount of heat needed to melt 24.0 kg of ice?
The amount of heat needed to melt 24.0 kg of ice can be affected by various factors such as the initial temperature of the ice, the surrounding temperature, and the pressure. The amount of heat required will also vary depending on the properties of the ice, such as its purity and crystalline structure.
4. Can the amount of heat needed to melt 24.0 kg of ice be reduced?
Yes, the amount of heat needed to melt 24.0 kg of ice can be reduced by increasing the surrounding temperature or by using a substance with a higher specific latent heat of fusion. This is because a higher temperature or a higher specific latent heat of fusion will require less heat to be transferred in order to melt the ice.
5. Is the amount of heat needed to melt 24.0 kg of ice the same as the amount of heat needed to freeze 24.0 kg of water?
No, the amount of heat needed to melt 24.0 kg of ice is not the same as the amount of heat needed to freeze 24.0 kg of water. This is because the process of melting and freezing involve different changes in temperature and require different amounts of heat to be transferred. Additionally, the specific latent heat of fusion for ice is different from the specific latent heat of fusion for water, which will also affect the amount of heat needed.
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