How many Btu's are required to change five pounds of ice at 20°F to water at 160°F?
a. 2,608 Btu’s
b. 1,390 Btu’s
c. 1,304 Btu’s
d. 141 Btu’s
I am trying ((160-20)x.5)x5 but not getting any of the possible choices. What am I missing?
Thanks for the help,
Chris
you are missing two things.
a) you have to heat ice to 32F first
b) you have to melt ice at 32F
c)then you have to change the temp of water from 32 to 160..
Heat ice from 20 to 32= 6BTUx5lb=30
Melt Ice at 32=144x5lb=720
Change temp of water from 32to160=128x5LB=640
30+720+640=1390
Correct?
I think I got it. Thanks for the help. Cleared a lot of things up.
ware do you add the 30 in
Hi Chris! It looks like you're missing one crucial step in your calculation. You've correctly calculated the temperature difference (140°F) and multiplied it by the mass (5 pounds), but you forgot to include the heat capacity of ice.
To convert ice at 20°F to water at 20°F, you need to account for the heat required to raise the temperature of the ice from -20°F to its melting point at 32°F.
So the correct calculation would be ((32- (-20)) x 0.5 x 5) + ((160-32) x 1 x 5).
Let's break it down:
First part: ((32- (-20)) x 0.5 x 5) = (52 x 0.5 x 5) = 130 Btu's
This calculates the heat needed to raise the temperature of the ice from -20°F to its melting point at 32°F.
Second part: ((160-32) x 1 x 5) = (128 x 1 x 5) = 640 Btu's
This calculates the heat needed to raise the temperature of the water from its melting point at 32°F to 160°F.
Now, adding these two parts together, we get:
130 Btu's + 640 Btu's = 770 Btu's
So none of the given answer choices match the correct value in this case. Better luck next time, and keep up the good work in your calculations! If you need any more help or have any more questions, feel free to ask.
To calculate the number of BTUs required to change ice at 20°F to water at 160°F, you need to take into account the various stages of the phase change and the specific heat capacities of ice and water.
Let's break down the calculation step by step:
1. The first step is to calculate the heat required to raise the temperature of 5 pounds of ice from 20°F to its melting point of 32°F.
Heat required = (mass of ice) * (specific heat capacity of ice) * (change in temperature)
Here, the mass of ice is 5 pounds, the specific heat capacity of ice is 0.5 Btu/lb°F, and the change in temperature is (32 - 20)°F = 12°F.
Heat required = 5 * 0.5 * 12 = 30 Btus
2. The second step is to calculate the heat required to change the ice at its melting point into water at the same temperature.
Heat required = (mass of ice) * (latent heat of fusion)
The latent heat of fusion for ice is approximately 144 Btu/lb.
Heat required = 5 * 144 = 720 Btus
3. The third step is to calculate the heat required to raise the temperature of the water from its melting point of 32°F to 160°F.
Heat required = (mass of water) * (specific heat capacity of water) * (change in temperature)
Since the ice has melted into water, the mass of the water will still be 5 pounds. The specific heat capacity of water is approximately 1 Btu/lb°F, and the change in temperature is (160 - 32)°F = 128°F.
Heat required = 5 * 1 * 128 = 640 Btus
4. Finally, add up all the calculated heat values to get the total heat required.
Total heat required = 30 + 720 + 640 = 1,390 Btus
Based on the above calculation, the correct answer to your question is (b) 1,390 Btu's.