Calculating Heat Required for Ice Cube Transformation

[Boozehound]

How much heat is required to change a 46.6 g ice cube from ice at -12.5°C to water at 53°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)

then i used Q=cmt and Q=mL

Q=(2090)(.0466)(12.5)=1217.425J
Q=(.0466)(33.5E4)=15611J
Q=(2010)(.0466)(53)=4964.298J

then when i add all the numbers up i get 21792.723J total but the answer is wrong. i don't know if I am using the formulas in the wrong order or what. I've done these problems in chemistry but its been a couple years since i took one of those courses. any help is appreciated.

 

[Kenny Lee]

You use C_water, not C_steam for the last calculation (C_water is 4180)

 

[m2003]

Hello,

Thank you for your question. It seems like you are on the right track with your calculations, but there are a few things that need to be adjusted.

First, when calculating the heat required for the ice cube to reach 0°C, you should use the specific heat of ice, not the specific heat of water. So the correct calculation would be Q = (2090)(0.0466)(12.5) = 1217.425 J.

Next, when calculating the heat required for the ice cube to melt, you should use the heat of fusion (L) instead of the specific heat. So the correct calculation would be Q = (0.0466)(334000) = 15556.4 J.

Finally, when calculating the heat required for the water to reach 53°C, you should use the specific heat of water, not the specific heat of steam. So the correct calculation would be Q = (2010)(0.0466)(53) = 4996.98 J.

When you add all of these values together, you should get a total heat of 21770.805 J, which is closer to the correct answer. The slight difference could be due to rounding errors.

I hope this helps. Keep in mind that when working with heat and temperature conversions, it's important to use the correct specific heat and heat of fusion values for the substances involved. Good luck with your calculations!