The diameter of the aluminum cylinder is measured to be: (4.935 plusminus 0.003)

Question

The diameter of the aluminum cylinder is measured to be: (4.935 plusminus 0.003) cm. The hanging mass (paint bucket and 5 kg) for a Mechanical Equivalent of Heat experiment is recorded as (5.2977 plusminus 0.0002) kg. The aluminum cylinder is turned 234 times. Calculate the work done cranking the aluminum cylinder. The mass of the aluminum cylinder is (207.8 plusminus 0.1)g. Once the cylinder is turned through the numbers of turns as described in Problem 1, the cylinder's change in temperature is (1 1.2 plusmius 0.6)" C. Determine the amount of heat that entered the cylinder. Based on the results of Problem 1 and Problem 2, determine the value for the Mechanical Equivalent of Heat as determined in the experiment described. How does this compare to the accepted value?

Explanation / Answer

formula for work done is

W= Fd

F = mg

W= mg d

=mgN pi D

=5.2977*9.81(234*pi*4.935*10^-2)

=1885.4237 J

different work

dW =F*d(d) +d*d(F)

=51.9704*pi*3*10^-5 +36.2788*2*10^-4*9.81

=0.0761 J

(2)

The amount of heat required to raise temperature is

Q = M*c*delt T

=207.8*0.897*11.2

=2087.642 J

differentiate Q

dQ =c*delt T*d(M) + M*c*d(del T)

= 0.897*11.2*0.1 +207.8*0.897*0.6

=112.84 J

(3)

from (1) and (2) parts

W/Q =1885.4/2087.6 =0.903

The theoretical value is 4.186 (J/cal) =4.186/4.2 (J/J) =0.9967

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