Good day fellow engineers,
I am currently studying to tackle my second 1rst class exam the applied thermodynamics.
I've been going through the sample exam questions listed in the ship's library and have some questions regarding Question 10
The answer listed says 43cm for the diameter and i keep arriving at a diameter of 4,3m
here is how im currently solving it can you please highlight my mistake.
Question 10: A solid metal sphere at a temperature of 15oC. is placed in 100 litres of fresh water at a temperature of 95oC. It was found that when equilibrium conditions prevailed, the diameter of the sphere had increased by 0.15%. Calculate the original diameter of the sphere.
note: specific heat capacity of water = 4.187 kJ(kgK)-1
note: specific heat capacity of sphere = 0.8876 kJ(kgK)-1
Linear expansion coefficient of sphere = 23(10-6)/K.
Relative density of sphere material at 15oC. = 2.56.
Ans. 43.07 cm. Ref. DOT10.
Increase in diameter = Alpha X original diameter X Delta T
0.0015Dinital = (23x10-6) X Di X Delta T
Delta T = 0.0015Di \ (23X10-6) Di (Di cancels out)
Delta T = 65.22oC
T2 = Delta T + T1
T2 = 65.22 + 15
T2 = 80.22oC
Msphere X specific heat capacity of sphere X DeltaT sphere = Mwater X Specific capacity Water X DeltaT water
Msphere X 0.8876 X (T2-15) = 100kg X 4.187 X (95 - T2)
Msphere X 0.8876 X (80.22-15) = 100kg X 4.187 X (95-80.22)
Msphere = 6188.386 \ 57.889272
Msphere = 106.9 kg
Mass sphere = Volume X Density
106.9kg = Vol X 2.56
106.9\2.56 = Vol
41.57596651m3 = Vol
Volsphere = 4\3piXr^3
(41.75796651 \ 4.18879)^1\3 = r
Do it Once, Do it right.