- #1
mlostrac
- 83
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Hello again,
I did a lab where ice was added to 150 mL of warm water and the temperature was observed and recorded. (The mass of the ice was not important; but after it all melted the total volume of the water in the cup was 218 mL)
Here's the question I have, "Assuming that the paper cup is a good insulator and that no heat energy was lost to the outside environment, estimate the heat of fusion of water. Note that part of the heat energy was absorbed by the melted ice to raise its temperature from 0°C
to the final temperature. "
How would I go about doing this? Is it just Q = mL = 0.150 kg x 333 KJ/kg ?
Here's what I tried:
If I find the heat lost by the water using the "Q=mc/\T" equation I get 20 kJ for "Q".
Then using Q=mL,
I get L = 20kJ/0.15kg =133 kJ/kg
Does that look right, based on what they are asking? Since the actual heat of fusion is 333 kJ/kg, my experimental data must be off?
I did a lab where ice was added to 150 mL of warm water and the temperature was observed and recorded. (The mass of the ice was not important; but after it all melted the total volume of the water in the cup was 218 mL)
Here's the question I have, "Assuming that the paper cup is a good insulator and that no heat energy was lost to the outside environment, estimate the heat of fusion of water. Note that part of the heat energy was absorbed by the melted ice to raise its temperature from 0°C
to the final temperature. "
How would I go about doing this? Is it just Q = mL = 0.150 kg x 333 KJ/kg ?
Here's what I tried:
If I find the heat lost by the water using the "Q=mc/\T" equation I get 20 kJ for "Q".
Then using Q=mL,
I get L = 20kJ/0.15kg =133 kJ/kg
Does that look right, based on what they are asking? Since the actual heat of fusion is 333 kJ/kg, my experimental data must be off?