Lab Quiz 10

The formula for Question 1 is given at the end of Question 1.

The cb and cw are the specific heats of the brass and water, respectively. Maybe this would look better - c_b andc_w.

The specific heats are constant and can be found on the Lab Manual; Lab 10.
For the brass and water:
c_b = 0.092
c_w = 1.000

xc wrote:DeltaQ=sqrt((Deltam/m)^2+(DeltaT/T)^2)*Q
Q= (mb cb + mw cw )(Tf – Ti )

m_b=200 c_b=.092 T=7.82
m_w=350 c_w=1
Deltam=.155 DeltaT=.396
Q=((200*.092)+(350*1))*(7.82*2)
Q=5761.776

Deltam=sqrt((.155)^2+(.155)^2)=.2192031
DeltaT=sqrt((.396)^2+(.396)^2)=.56002857
m=mb cb + mw cw
m=(200*.092)+(350*1)=368.4
T=given T*2=7.82*2=15.64

Delta Q=sqrt((.2192031/368.4)^2+(.56002857/15.64)^2)*5761.776

I don't understand how you found Deltam and DeltaT. How can you use Deltam to find Deltam? Can you or someone else please clarify this?

what is cb and cw??

periwinkle - the errors for the mass and temperature are given to you in the question, you just have to plug it in.

noname - try checking the top of the page.

1) You do work against a constant force of F = 2.78 N, which balances (is equal to in magnitude) the frictional force between the two brass cups shown, the rotating outer brass cup and the stationary inner brass cup in the sketch above. The force exerted by a string acts tangentially to the disk shown, which has an effective diameter D = 10.4 cm. What is the amount of work (Ch8 sheet 38) you do via a crank that rotates the outer brass cup N = 2397 times ? (Hint: when evaluating the work, write down the torque given by the tangential force and the lever arm, which is the effective radius of the disk. Note that each turn of the outer brass cup corresponds to an angle of 2 p radians) Indicate with a negative (positive) sign whether the equation for the work, W = N F 2p D/2, is correct and follows ( is incorrect and does not follow) from the expression in Ch8 sheet 38.

For question 1....
I know the formula is given but what do you plug in for 2p?

You do work against a constant force of F = 2.8 N, which balances (is equal to in magnitude) the frictional force between the two brass cups shown, the rotating outer brass cup and the stationary inner brass cup in the sketch above. The force exerted by a string acts tangentially to the disk shown, which has an effective diameter D = 10.3 cm. What is the amount of work (Ch8 sheet 38) you do via a crank that rotates the outer brass cup N = 2,174 times? (Hint: when evaluating the work, write down the torque given by the tangential force and the lever arm, which is the effective radius of the disk. Note that each turn of the outer brass cup corresponds to an angle of 2 p radians) Indicate with a negative (positive) sign whether the equation for the work, W = N F 2p D/2, is correct and follows ( is incorrect and does not follow) from the expression in Ch8 sheet 38.

Thank you

It's not 2p, its supposed to be 2pi, as in pi=3.14

I know this is probably annoying everyone but I've been trying to do #3 countless of times using the forumla provided and for some reason mapleTA constantly marks me incorrect. I don't know what I'm doing wrong. This is what I did for one of my prevous attempts:

This was the question:
When doing the experiment sketched above you heat up the brass containers with a combined mass mb = 200 grams and an amount of water with a mass mw = 350 grams. What is the absolute error of the amount of heat absorbed (Ch 13 sheet by the cups and the water (in calories), if their initial temperature (at the start of the cranking) and their final temperature (after ending the cranking) is 7.63 0C below and above room temperature, respectively (Use the specific heats given in Table 1 of the lab manual for this experiment 10). Assume that the error of the brass and water masses is 0.15 grams and the error of the initial and final temperature is 0.472 0C each.

This was my work:
mb = 200 grams cb = .092
mw = 350 grams cw = 1
T = 7.63
Deltam = 0.15 grams
Delta T = 0.472 0C
Q = ((200 x .092)+(350 x 1)) x (7.63 x 2)
Q = 5621.784

Deltam = sqrt((0.15)^2 + (0.15)^2) = .212132
Deltat = sqrt((0.472)^2 + (0.472)^2) = .667508

m = (200 x .092) + (350) = 368.4
T = givenT x 2 = 7.63 x 2 = 15.26
Delta Q = sqrt((.212132/368.4)^2 + (.667508/15.26)^2) x 5621.784 = 245.9299 (positive)


Can someone please tell me what I did wrong? Thank you so much in advance!