A Virtual Heat Lab: Webassign is using Caluminum 90 g deg. Project 1 An aluminum
ID: 306709 • Letter: A
Question
A Virtual Heat Lab: Webassign is using Caluminum 90 g deg. Project 1 An aluminum cup has a mass of 69.0 grams. You put 116.0 mL of tap water into the cup: the cup and the tap water are both at 23.0°C. You take 108.0 grams of aluminum pellets and heat them over a steam pot until they reach 94.0°C. Then you dump the pellets into the cup of tap water. After a while, you measure the temperature of the water in the cup and find that it reaches 30.4°C How many joules of heat did the pellets give off? 6182 How may joules of heat did the aluminum cup absorb? 460 How may joules of heat did the tap water absorb? 3593 How many joules of heat were lost to the surroundings? 2129 Project 2 So you take the same cup of tap water starting at the same temperature 23.0°C 1. J. This time you use a rubber tube to pipe hot steam into the Cup until the temperature inside the cup reaches 60 When you are done, the volume of water in the cup has increased by 15.0 mL. How many joules of heat did the metal cup absorb? How many joules of heat did the tap water absorb? How many joules of heat did the additional 15.0 mL of 100°C water give off as it cooled down to 76.6°C? This doesn't add up: so how many joules of heat should you figure were given off by the 15.0 g of steam condensing back into 100°C water? Based on your results, what is the latent heat of vaporization in Joules per gram? J. J/g.Explanation / Answer
Heat absorb by cup = mcup*cAl*?T
?Qcup= mcup*cAl*(Tf – Ti)
Plugging values,
?Qcup= 0.069*900*(76.6-23.0)
?Qcup= 3328.56 J
Heat absorb by cup = mtap water*cAl*?T
?Qtap water = mtap water*ctap water*(Tf – Ti)
Plugging values,
?Qtap water= (0.116+0.015)*4181*(76.6-23.0)
?Qtap water= 29357.31 J
Heat absorb by cup = mwater*cAl*?T
?Qwater= mwater*cwater*(Tf – Ti)
Plugging values,
?Qwater= 0.015*4181*(76.6-100.0)
?Qwater= -1467.53 J
Heat given of by 15g water into 100o water = ?Qcup+ ?Qtap water + ?Qwater= 29357.31+3328.56 -1467.53 = 31218.34 J
?Qgiven off = 31218.34 J
m*Lf = ?Qgiven off
Lf = (?Qgiven off)/m = 31218.34/0.015 = 2.08*10^6 J/kg
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