HW14b help and hint:
Topic: Homework 14a #1
Student A: For question number 1 i am given a diagram/ graph showing pressure vs Q(heat
added) and it asks me for the time it takes to reach each point(freezing to melting, melting to
boiling). first of all, how do i find Q at these points, and then how do i get time using Q and
P?
Man, Weining: First, the ice need to raise T to melting T,
heat needed is mc(deltaT), where c is the specific heat of ice...
Then during melting the T doesn't raise, but heat is absorbed for melting right? latent heat for
fusion per kg of ice can be found on text book. Lf
Q=m*Lf
After it is all melted, the WATER needs to increase T and absorb Q=mc(deltaT), here c is the
specific heat of water....
Power = work/time
also power = heat/time.
If you find how much Q you need, you can find time needed at given power.
If you know time, you know how much Q is added and you can find out what has happened to
the ice...
Student B: I am having trouble using the heat that I find and trying to find time using power
and heat. When I plug it into Power = heat/time it does not give me the right answer. What am
I doing wrong?
Student C: What number do i use for power?
Man, Weining: The figure shows the temperature of 1.000 kg of water as heat is added to the
system. Suppose that such a sample of water starts at point A at time zero. Heat is added to
this system at the rate of 13750 J/s. Calculate the time for the system to reach each of the
following points.
First, you need to find out to reach each point, A, B, or C what is the total Q needed. You
need to calculate the Q needed to reach those points accurately with correct unit (J).
Once you know the Q needed to be added, you can find time because they told you, the Q was
added at rate of 13750 J/s.
13750 J per second is the power for that heating system. Meaning for every second, 13750J of
heat is added to the system.
If the Q needed to reach point A is 1375000 J, the time needed is simply Q/Power = 137500J /
(13750J/S) = 100 seconds.
#3
A sample of pure water is initially at atmospheric pressure and has a temperature that is just
above the boiling point.
(a) If the temperature of the sample is increased while the pressure is held constant, what
phase changes occur?
(b) Suppose, instead, that the pressure of the sample is decreased while the temperature is held
constant. What phase changes occur now?
Weining :
starting at a point at atmospheric pressure and has a temperature that is just above the boiling point
keep p constant (horizontal) move rightward (T increase), does the phase change? which kind?
Does it across any boundaries between two phases? From what phase to what phase?
starting at a point at atmospheric pressure and has a temperature that is just above the boiling point
keep T constant (vertical) move downward (P decreas), does the phase change? which kind?
Does it across any boundaries between two phases? From what phase to what phase?
Topic: Homework 14a#4
A 5.0 kg block of ice at -1.5°C slides on a horizontal surface with a coefficient of kinetic
friction equal to 0.060. The initial speed of the block is 6.3 m/s and its final speed is 5.5 m/s.
Assuming that all the energy dissipated by kinetic friction goes into melting a small mass m of
the ice, and that the rest of the ice block remains at -1.5°C, determine the value of m.
Student A: I probably need to review friction again but i am having trouble figuring out this
question. I answer the amount of friction force caused by using Uk*N(or mg). My problem is
finding out the work friction has done to the ice block. Since i only have Vf and Vi i don't
know the amount of time or distance that has travelled or past. So w/o acceleration, distance,
or time how can i find work?
Man, Weining: Right, that's exactly what you need to review:
Finding out the work friction has done to the ice block.
Work of friction = friction force * distance *(-1)
If you can find distance, you can find how many J of work is done by friction. (that many J of
heat is created then).
How to find distance? analyze the forces, what is total force on the object? what would be
acceleration? is that constant acceleration? if yes, with a, vf and vi, can you find
displacement....
Also there is a shortcut for this problem. In stead of finding the distance and Work of friction
= friction force * distance *(-1). Do you know what kind of consequence the work did to the
total energy E of the object?
Ef= Ei+Wnc If you know how much was the energy (total KE and PE) change, you know how
much work was done Wnc=Wfriction here, right?.
So, can you easily find the work done by friction by looking at mechanical Energy change?
(question 1b in midterm 2).
Student B: I am having a difficult time with this problem even after to talking to you, so I
know I am missing something important.
The only thing left is to think that Wnc=1/2mvi^2-1/2mvf^2, using the initial mass in the
equation for initial and final and calculating the difference. At least that way I get a total
work done that I can further calculate, but this does not take into account the mass that is lost
by friction, which should decrease final KE somewhat. What am I missing?
Man, Weining: We don’t consider “mass lost by friction”. Friction doesn’t reduce mass.
Friction takes away some kinetic energy and converted it into heat. Small amount of ice
absorbed that heat and increased temperature from -1.5 C to 0C and then melt. That amount is
so small comparing to the 5kg block. So when we find work done by friction by
Wnc=1/2mvi^2-1/2mvf^2, we still use 5kg for the final mass. Once we get the work done by
friction, it is equal to the total heat absorbed by the small amount of melted ice. (m_melt). You
will need specific heat of ice c_ice = 2030 J per kg per degree.
And latent heat for fusion: Lf = 33.5 *10^4 J/kg
(you see. Lf is such a larger number, that the heat generated by friction can only melt a small
amount of ice)
Topic: Homework 14a #5 An ideal gas is taken through the three processes shown in figure
below.
Student A: I understand that change in U=Q-W. But if we are given only 1 value out of these
three how are we supposed to find the other 2?
Weining: It is important to notice when Volume of gas do not change, Work done by the gas
is zero.
It is also important to notice that from state 1 to 2 to 3 back to 1 total change in internal
thermal energy is deltaU(1 to 2) +deltaU(2 to 3)+ deltaU(3 to 1) and that is equal to zero,
because after 3 steps it goes back to initial P,V and T.
Topic: Homework14a#7
Consider a system in which 4.80 mol of an ideal monatomic gas is expanded at a constant
pressure of 101 kPa from an initial volume of 2.15 L to a final volume of 4.19 L.
(b) Find the change in temperature for this process.
Weining: you know not only initial and final values of p and V, you also know the number of
moles of gas (the amount of gas)
As a result using PV/T = nR, you can find out T. You can do that for both initial T and final T
separately.
Student A: I got the answer right, but I have a question about the units for pressure. The
problem states pressure in kPa. When I converted it to Pa in calculations, the answer was
wrong, but when I kept it in kPa, I got the right answer. Should we keep the pressure in units
of kPa for these calculations?
Man, Weining: That was a lucky coincidence. I am glad that you pointed it out.
For pressure, 1 kPa is 1000 Pa
For volume, 1L (1 litter) is 0.001 meter cubic.
1Litter is 10cm by 10 cm by 10 cm, it’s 0.1 m by 0.1 m by 0.1m, (a cubic with the side edge
as wide as your hand). 1 L is only 0.001 meter cubic.
You didn’t convert L to meter cubic (your volume is off by a thousand time) and you also
didn’t convert kPa to Pa, so your Pressure is off by a thousand times, (1/1000)
Pressure multiply by Volume, you happened to end up with the right numbers.
It’s a lucky coincidence and you really want to use right units and understanding them.
Student A: Thanks Professor Man. I noticed units conversions are not on the final equation
sheet online. Will they be given on the exam?
Man, Weining: For things as common as kilo = 1000,
1km =1000m
1kg =1000g
1m = 1000mm,
1cm =10mm
1 L = 1000 ml (1 litter = 1000 ml)
etc, you should have known them by now. (if not sure ask me during the exam, is fine. )
Other things like 1 inch =2.54 cm
1 meter cubic = 1000 L.
1mile =1600m would be given on the test, if needed.