please show me detail works, thanks The diameter of the aluminum cylinder is mea
ID: 2302880 • Letter: P
Question
please show me detail works, thanks
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 plusminus0.0002) kg. The aluminum cylinder is turned 234 times. Calculate ilic work done cranking the aluminum cylinder. The mass of the aluminum cylinder is (207.8 plusminus 0.l)g. Once the cylinder is turned through the numbers of turns as described in Problem 1, the cylinder's change in temperature is (l 1.2 plusminus 0.6)degree 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
1) Work = Force*displacement
F =M*g =5.2977*9.81 =51.9704 N
d =N*pi*D =234*pi*4.935*10^-2 =36.2788 m
W =F*d =1885.4237 J
Error is (we differentiate W):
dW =F*d(d) +d*d(F) =51.9704*pi*3*10^-5 +36.2788*2*10^-4*9.81 =0.0761 J
2)
caloric capacity of Aluminum is
c =0.897 J/g/K (wiki)
Total heat is
Q = M*c*delta(T) =207.8*0.897*11.2 =2087.642 J
Error is (we differentiate Q):
dQ =c*delta(T)*d(M) + M*c*d[delta(T)] = 0.897*11.2*0.1 +207.8*0.897*0.6 =112.84 J
3)
Mechanical equivalent of heat is the raport between the mechanical work and the heat generated
Equivalent = W/Q =1885.4/2087.6 =0.903 (J/J)
The theoretical value is about 4.186 (J/cal) =4.186/4.2 (J/J) =0.9967 (J/J)
(http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat)
(Observation 1 calorie =4.17-4.22 J is not fixed and depends on the temperature)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.