NEED EXPLANATIONS; EQUATIONS ON A PAPER OR HERE Question 7) You draw a 10 kilogr

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NEED EXPLANATIONS; EQUATIONS ON A PAPER OR HERE

Question 7)
You draw a 10 kilograms of water up ten meters from a well. Approximately how much work energy did you provide? (Remember that you are doing work as force times distance. You will want to convert the 10 kg into force of the gravity against which you are working.)
a) 1 Joule b) 100 Joules c) 1 kiloJoule d) 10 kJ e) 100 kJ

Question 8)
You heat that same 10 kilograms from 20 oC to 30 oC. Approximately, how much heat energy did you supply to the water? (Recall, Table 5.3, that the heat capacity of water is 4,180 Joules/(kg oC).)
a) 400 Joule b) 4 kiloJoule c) 40 kiloJoule d) 400 kJ e) 4000 kJ

Question 9)
In class you were told that it takes 334,000 Joules, the "latent heat of fusion," to melt 1 kg of ice. How much heat must one remove from 10 kg of water to freeze it into ice?
a) 33.4 kiloJoules b) 334 kJ c) 3340 kJ d) 334,000 kJ

Question 10)
You were also told that it takes 2,260,000 Joules (2,260 kJ), the "latent heat of vaporization," to boil a kilogram of water. How many times more heat does it take to boil vs. to melt a given amount of water?
a) 3.5 times b) 7 times c) 35 times d) 70 times e) 350 times

Question 11)
In class we discussed the perfect gas law, particularly the relation (pressure x volume)/temperature equals a constant, pV/T = constant. If you compress a gas and maintain its temperature, you make its volume smaller. If you do this and you compress the gas to half its volume, by how much must you increase its pressure?
a) 25% b) 50% c) 100% d) 200 % e) 400%

Question 12)
A heat engine is a device that takes heat from a "source" at one temperature Thot, extracts some of the energy of that heat to make mechanical work, then dumps the remainder of the energy into a "sink" at a lower temperature Tcold. As described in class, the "efficiency" of such a device is limited by a strict rule that the work extracted cannot be more than (1

Explanation / Answer

7) Work=M*g*h =gain in potemtial energy=10*10*10=1 KJ Hence C

8) 4,180 *10*10=400KJ Hence C

9) 33400 *!0=3340 KJ Hence C

10) 260,000/33400=7 times Hence B

11) The pressure becomes 2 times.Hence 100% increase Hence C

12) effeciency =1 -(273/373)=0.27 Therefore Effeciency= 25% Hence A

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