- #1

FestiveF

**Thermal Energy/ States of Matter- Help!!**

Hey guys! I have been on the road traveling for 6 weeks straight with not much time for schoolwork. Now I am home for 4 days so I have to get to work. I have finished one of my physics tests but wanted some reassurance that the answers are correct- as I cannot afford a bad grade- must keep the grades up. I will post the problems and my solutions and if you could check over them that would be greatly appreciated! Thanks ahead of time!!!!!

**1) Which takes less time to melt, one Kg of ice, one Kg of lead, one Kg of copper, or one Kg of silver? Assume that ea h solid is at its melting point and the heat energy is being applied to each solid at the same rate.**

I beleieve that the answer to this one is silver because it has the lowest heat of vaporization at 1.04 x 10^4.

**2) Container A has 50 Kg of water in it. container B has 30 Kg of water in it. If both containers have equal quantities of heat energy then__________.**

I beleieve that container B and its contents would have a higher temperature than conatiner A due to the fact that there is less material within it. Is this right??

**3) If the density of mercury is 13.59 x 10^3 kg/m^3 at 20 * C, what will its density be at 65 * C?**

OK...this is where I am confused. I believe that you somehow have to incorporate the equation Vi(coefficient of volume expansion)(change in temperature)= m^3....so you would use 13.59 x 10^3 (180 x 10^-6)(65*-20*). Does this sound correct? I got the answer to be 675 N but that doesn't seem correct....

**4) In a certain hrdraulic liff, the small piston has a radius of 15 cm. The large piston has a radis of 26 cm. A force of 225 N is applied to the small piston. What is the mass of the crate that is being lited by the application of this 225 N force?**

Here I know that you have to use Pascal's principle and the equations P1= F1/A1...(pressure= force/surface area)and F1/A1=F2/A2....I don't understand this enough to solve it...

**5) The heat of vaporization of mercury is 2.72 x 10^5 J/Kg. How many grams of mercury at its boiling point can be vaporized by the addition of 4.53 kJ of heat energy?**

Here, I think you would use Q=m(Hv)

4530 J= m(2,72 x 10^5)

m= .01665 kg or 16.7 g

Thank you so much for your time- I really appreciate it!!!