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.

Solution

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

r= 2.15M

D= RX2

D=4.3M