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Q: Hard Physics Questions ( Answered 5 out of 5 stars,   2 Comments )
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:5 out of 5 stars
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:

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:5 out of 5 stars and gave an additional tip of: $1.00

Subject: Re: Hard Physics Questions
From: mclean48-ga on 10 Nov 2004 12:00 PST
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)


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.
Subject: Re: Hard Physics Questions
From: veky-ga on 12 Nov 2004 19:13 PST
//* it is interesting to think that you can replace the baseball with anything. *//

You mean, anything with the same specific heat.

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