HW13a part 1 help and hint:
Topic: Homework13b#2
The gas in a constant-volume gas thermometer has a pressure of 102.0 kPa at 99°C.
(a) What is the pressure of the gas at 50.0°C?
(b) At what temperature does the gas have a pressure of 115 kPa?
Student A: Not sure if i am doing this correctly but i am using the equation p/t=constant
volume in order to answer questions a and b but i am not getting the right answer. The
question seemed pretty easy but i don't understand what i am doing incorrectly.
Weining: Yes. P/T doesn’t change, initial = final. But T should be in Kelvin instead of C.
Pi/Ti = Pf/Tf
: also
Ti/Tf = Pi /Pf. Notice that the temperature ratio Ti/Tf is very
different in Kelvin units or degree C units. The ratio is only useful in Kelvin unit.
Topic: #5, 6, 7
Man, Weining: Please read the last chapter. chapter 18 over the weekend.
You need to read chapter 18 in order to answer the last 3 conceptual questions in HW
13b.
The last 3 questions are reading assignment.
13a part 2 help and hint
Topic: Homework 13a #1
One day you notice that the outside temperature increased by 27°F between your early
morning jog and your lunch at noon. What is the corresponding change in temperature in
each of the following scales?
Student A: I converted the outside temperature increase in celsius. I thought the
conversion to kelvin was simply the Tc + 273.15. After i plugged in the answer for some
reason it says that i am incorrect am i missing a step because this seems very easy and
for some reason i am incorrect.
Man, Weining: Good, also For part 1a, you need to pay attention:
I emphasized this 3 times today during the class. :)
The change of T is not T itself.
When you increase T in Tc for 5 Degree C, how much is increased in terms of T in
degree of F?
The concept of SLOPE!
TF= TC *9/5+ 32
question 1a asks you to Find change of TC when you know change in TF.
Find the corresponding CHANGE.
suppose your initial T was TF=0, now it is TF=27 F
what was initial in TC and final in TC?
Find the change. And you need to understand the concept of slope in this equation:
TF= TC *9/5+ 32
How is change in TF related with change in TC?
How is Change in Kelvin related with change in TC?
Student B: Im using equation Tc= 5/9(Tf-32) and it is still not giving me the right answer
and to get Kelvin Im adding 273 to that answer and it says im off by orders of magnitude.
Am I doing something wrong?
Student C: They are not asking you to convert 30 degrees F to Celsius or Kelvin. They
are asking you by how much the degrees in Celsius/ Kelvin changed.
Remember what Man said? For every 5 degrees Celsius you get 9 degrees Fahrenheit.
So if the temperature increased by 30 degrees Fahrenheit. How much degrees Celsius
and Kelvin increased?
i.e. Say you started at 32 degrees F; now its 62 degrees F..
but in Celsius its 0 degrees ,and if it increased 30 degrees F, how much did it really
increase in Celsius?
So hard to explain clearly without giving away too much information. :P
Student D: So if my equation is right why am i getting the wrong answer?
Man, Weining: That equation from student B is not right,
You need to understand the question first.
you were asked to find the CHANGE of T in degree C, not T in degree C.
Also i have posted the explaination in the HW discussion forum in the 13A 1a thread.
There are two ways to say a baby's age:
by months or by years.
7 month baby or 17 month baby or 1.3 year old baby...
now if the baby's age INCREASE by 15 month, how did his age INCEASE in terms of
years?
12 month of increase = 1year increase
15 month of increase = 15/12 year of increase
so the increase is 1.25 years
if you plot baby age in month in y axis and baby age in years in x axis, what matters is
the SLOPE.
one step increase in years = 12 steps increase in months.
I emphasized this during the class. :)
the change of T is not T.
When you increase T in Tc for 5 Degree C, how much is increased in terms of Tf in
degree of F? The concept of SLOPE!
TF= TC *9/5+ 32
Find change of TC when you know change in TF.
(similar to finding the change of baby age in months vs baby age in years, only slope
matters)
Find the corresponding CHANGE.
suppose your initial T was TF=0, now it is TF=23 F
what was initial in TC and final in TC????
Find the change. And you need to understand the concept of slope in this equation:
TF= TC *9/5+ 32
Topic: Homework 13a, #4
Early in the morning, when the temperature is 5.0°C, gasoline is pumped into a car's 61
L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27°C.
Since the volume of gasoline increases more for a given temperature increase than the
volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out
in this case?
Student A: It says I am within 10% range for the right answer but I dont know what else
I can do because all the numbers are whole numbers except for the coefficient of volume
expansion for gasoline...the coefficient is 0.00095, the volume is 61L, and the change in
temp is 22 degrees. I cant get more significant figures if I wanted to..
Student B: you need also need to think about how the gas tank made of steel expands and
take that into account as well. its like example 16-3 on page 548 in the book. i also had
trouble with this at first as well
Weining: change of vol of steel tank: should be 3*alpha*Volume*change of T
Why is there a factor of 3?
When the length increase by 1 %
from 1 meter to 1.01 meter
The volume will increase from 1meter cubic to 1.01^3 meter cunib
1.01^3 =1.03 so the volume increases 3%
Topic: Homework 13a #6
Water, iron all have the same final Temp Tf
water initial Ti_water
Iron initial Ti_Iorn
the total Q =0.
Q=mc*(Tf-Ti)
; Q_water+ Q_iron =0
Or
The total Q absorbed (absolute value) = the total Q gave off (absolute value)
Who gave off heat? the Iron; Who absorbed heat? the water
The most common mistake is to mess up with positive or negative values.
Initially before mixing, there was a Hotter Temp, and a Colder Temp.
You know that the Final temp Tf for all objects is equal after mixing.
Tf is less than the initial hot Temp, and Tf is higher than the initial cold Temp.
If you always use Tf minus Ti to calculate Q, you get some positive values of Q and
some negative value of Q, add them together, it should be zero.
Or, If you consider the absolute value of Q_absorbed and Q_gaveoff, and set them equal
to each other, you mean two positive values should be equal. In that case you should
always use the higher Temp minus the lower Temp, so that you get the absolute value
(postive) for both gave off and absorbed heat.
Also, since the only thing matters here is Temperature difference: Tf-Ti, no matter you
use degree C or use Kelvin, the result is the same.
Topic: Homework 13a #7
To determine the specific heat of an object, a student heats it to 100°C in boiling water.
She then places the 33.9 g object in a 155 g aluminum calorimeter containing 89 g of
water. The aluminum and water are initially at a temperature of 20.0°C, and are
thermally insulated from their surroundings. If the final temperature is 22.0°C, what is
the specific heat of the object?
Student A: Do I need to take into account the Q of the aluminum?
Man, Weining: Do I need to take into account the Q of the aluminum?
Yes. That's why they gave you the mass of the container.
Water, aluminum, and the object all have the same final Temp Tf=22C
water initial T= 20 C
Aluminum initial T =20C
the object initial T was 100C.
Consider the three things together, the total Q =0.
or
The total Q absorbed (absolute value) = the total Q gave off (absolute value)
Who did give off heat? the object
WHo absorbed heat? both the water and the ALuminum container...
Q=m c (Tf -Ti) for each different object.