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course Phy 202
1/19 3
Set 54 Problem number 2Problem
Water is filled to a depth of 4 meters above a small hole in a large container. Use energy conservation to determine the velocity of 7 grams of water as it flows out through the hole, provided there are no dissipative losses.
a=fnf(2,20,1) b=fnf(3,9,1) c=b/1000 d=c*9.8 e=d*a f=fng(2*e/c)
I have completely no idea how to do this one to be honest! Im going to start by looking at the general solution and see if i can figure it out how to do it.
PE = -mgy 0.007kg *9.8m/s^2*-4m =-0.2744 J
This is as far as I can get on this problem. Would you mind please going over this on Monday?
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Sure, but since we don't have class tomorrow it will have to be Wednesday.
In the meantime consider the following:
PE decreases by .27 Joules so KE increases by .27 Joules.
Initial KE was zero, as the water was originally stationary.
So the final KE must be .27 Joules.
Set 1/2 mv^2 equal to .27 Joules and solve for v.
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Set 54 Problem number 10
Problem
A wall has area 11 m^2 and constant thickness .24 m. The inside of the wall is at 193 Celsius while the outside is at 3 Celsius. What is the temperature gradient from inside to outside?
190 C between 0.24m=24cm -7.917C/cm
If the outside temperature increased, thereby increasing the temperature difference by a factor of 1.254, what would be the temperature gradient?
-7.917(1.254) = -9.928C/cm
Set 54 Problem number 12
Problem
By measuring the temperature changes of substances with which a sample comes into contact, we find that that 60.84 Joules of energy are required to increase the temperature of a 13 kg sample of a certain substance by 12 degrees Celsius.
• What is the specific heat of the substance?
For some reason, I feel like I don't have enough information to answer this. I know I do, I just don't know where to put the information I have.
I was trying to use PV=nRT which is why I though I needed all the information.
ThermE per kg =4.68J/kg /12C= 0.39J/kgC
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The given solution explains what a specific heat is. Once you know that it's possible to attempt the problem.
The specific heat is the energy required to raise 1 gram, or if you prefer 1 kilogram, of the substance by 1 degree Celsius. The units of the specific heat will be Joules / ( kg * Celsius) or Joules / (gram * Celsius).
How many Joules did it take to raise 13 kg by 12 Celsius?
How many Joules is this per kilogram (or if you prefer, per gram)?
How many Joules per kilogram (or Joules per gram) are required for every Celsius degree increase in temperature?
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