Lab Quiz 10

1) 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 = 10.6 cm. What is the amount of work (Ch8 sheet 38) you do via a crank that rotates the outer brass cup N = 1852 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.


--------------------------------------------------------------------------------




2) When doing the experiment sketched above you heat up the brass containers with a combined mass mb = 168 grams and an amount of water with a mass mw = 339 grams. What is 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 4.92 0C below and above room temperature, respectively (Use the specific heats given in Table 1 of the lab manual for this experiment 10) ? Indicate with a negative (positive) sign whether the equivalent amount quoted in SI units is different from (equal to) your result in calories.


--------------------------------------------------------------------------------




3) 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 4.150C 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 0.186 grams and the error of the initial and final temperature is 0.623 0C each. (Hint: In Q= (mb cb + mw cw )(Tf – Ti ) calculate the error of each bracket according to expression (6) in “Error and Uncertainty” (“EU”). Note that cb and cw are factors which must be included in the absolute errors when applying (6). Then calculate the error of the product of the two brackets according to expression (7) of “EU”). Indicate with a positive (negative) sign whether you can (cannot) replace the 0C by the same number of 0K.


Please help...thank you

they give you the formula so just plug in the #'s
answer is neg according to the CD

#2 is also just plugging in the number to the equation given, and the answer is negative
anyone figured out #3?

I also need help with number 3 did anyone figure it out?

There is no equation posted with the problem. It tells you to refer to the chapter, but it is not very clear on how to do the problem.....

can someone just put up some explanations thatd be sickkkk and much appreciated...thanks

Pleaseeeeeeee
help me with #2 and #3

i finally figured out 2

Q=(mbcb+mwcw)(change in temp)

change in temp is just the # in degrees they gave multiplied by 2

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

is 3 negatgive? and do we convert grams to kilograms for number 2? thanks

can someone explain 3 plz?

use grams, specific heat c is used as cal*g-1*C-1

For 1, the formula given in the question is correct, so answer is negative. The method for number two above is correct, also negative. Number three solution above is also correct, answer is positive.

I still cannot get #3. it is not working on mapleTa and the question says that the specific heats are suppose to be in the error equation.

All the formulas have worked for me, can we see your work?

don't have the cd...can you please write the formula to 1. Also. what does cb cw stand for?