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Q: Hard Physics Questions ( Answered ,   2 Comments ) Question
 Subject: Hard Physics Questions Category: Science > Math Asked by: scifinut69-ga List Price: \$15.00 Posted: 07 Nov 2004 00:51 PST Expires: 07 Dec 2004 00:51 PST Question ID: 425619
 ```A 4 kilogram block of ice is removed from a freezer where its temperature was maintained at ? 20 degrees Celsius. How much heat does the ice absorb as it is warmed to ? 10 degrees? (The specific heat capacity of ice is 2,000 joules per kilogram degree Celsius.) The Eiffel tower in Paris is 300 meters tall on a cold day (T = ? 5 degrees Celsius). What is its height on a hot day when the temperature is 35 degrees? (It is made of iron which has a coefficient of linear expansion of 0.000012 per degree Celsius.) A baseball with mass 0.14 kilogram and speed 40 m/s is caught. If all of its kinetic energy is converted to heat as it is caught, and all of the heat is absorbed by the ball, what is its temperature change? (Assume the baseball?s specific heat capacity is 1,000 joules per kilogram degree Celsius.) On a winter day the temperature is 5 degrees Celsius and the humidity is 0.003 kilograms per cubic meter. a) What is the relative humidity? b) The air is brought into a building and heated to 20 degrees Celsius without changing the humidity. What is the relative humidity inside the building?``` Subject: Re: Hard Physics Questions Answered By: redhoss-ga on 07 Nov 2004 08:08 PST Rated: ```Hello scifinut69, I think these are your answers: 1. dQ = m x c x dT dQ = 4kg x 2000 J/kg C x 10 C = 80,000 Joules 2. dL = coef linear exp x dT x L initial dL = .000012 x 40 C x 300 m = .144 meter 3. KE = .5 x M x V^2 KE = .5 x .14kg x 1600 m^2/sec^2 = 112 kg m^2/sec^2 = 112 joules dT = dQ / m x c = 112 joules / 1000 x .14 = .8 degree Celcius 4. I had to work a little on this one. You have to know what the saturated capacity of air is at your stated temps. Here is the link to that info: http://cougar.slvhs.slv.k12.ca.us/~pboomer/grades/demos/atmos.html Now it is easy: (a). RH = 3/8 = 37.5% (b). RH = 3/18 = 16.6% NOTE: .003kg = 3 grams Hope this helps you out, Redhoss```
 scifinut69-ga rated this answer: and gave an additional tip of: \$1.00 ```An interesting point to note about "number 3" - the baseball, heat problem is that the mass of the baseball is irrelevant. Using Energy in = Energy out: (1/2)*(m)*(dv)^2= (dT)*(C)*(m) a little algebra and we get: dT = (dv)^2/(2 * C) where; dT = change in temp. dv = change in velocity C = specific heat Although this problem is very ideal (all energy being retained in baseball, heat is absorbed uniformally thoughout baseball), it is interesting to think that you can replace the baseball with anything.```
 ```//* it is interesting to think that you can replace the baseball with anything. *// You mean, anything with the same specific heat.``` 