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T-Mobile 7:30 PM a webassign.net A Virtual Heat Lab Webassign is using Caluminum

ID: 306696 • Letter: T

Question

T-Mobile 7:30 PM a webassign.net A Virtual Heat Lab Webassign is using Caluminum 900 g deg- Project 1 An aluminum cup has a mass of 69.0 grams. You put 119.0 mL of tap water into the cup; the cup and the tap water are both at 20.0°C. You take99.0 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 28.4°C. How many joules of heat did the pellets give off?3 How may joules of heat did the aluminum cup absorb? How may joules of heat did the tap water absorb? How many joules of heat were lost to the surroundings?J. Project 2 So you take the same cup of tap water starting at the same temperature (20.0°C). This time, you use a rubber tube to pipe hot steam into the cup until the temperature inside the cup reaches 62.5°C. When you are done, the volume of water in the cup has increased by 11.0 mL. How many joules of heat did the metal cup absorb?1 How many joules of heat did the tap water absorb? How many joules of heat did the additional 11.0 mL of 100°C water give off as it cooled down to 62.5°C? This doesn't add up: so how many joules of heat should you figure were given off by the 11.0 g of steam condensing back into 100°C water? J. Based on your results, what is the latent heat of vaporization in Joules per gram?

Explanation / Answer

1.

Heat given off by pellets = mpellets*cAl*?T

Qpellets= mpellets*cAl*(Tf – Ti)

Plugging values,

Qpellets= 0.099*900*(28.4-94.0)

Qpellets= - 5844.96 J

Heat absorb by cup = mcup*cAl*?T

Qcup= mcup*cAl*(Tf – Ti)

Plugging values,

Qcup= 0.069*900*(28.4-20.0)

Qcup= 521.64 J

Heat absorb by cup = m*cAl*?T

Qwater= mwater*cwater*(Tf – Ti)

Plugging values,

Qwater= 0.119*4181*(28.4-20.0)

Qwater= 4179.34 J

Heat loss = Heat given off by pellets – (Heat absorbed by cup+water)

Heat loss = 5844.96 - (4179.34+521.64)

Heat loss = 1143.98 J