A closed vessel made of brass 4mm thick ,has a surface area 0•2m^2 and contains 5×10^-3m^3 of hot water.calculate the Temperature difference which exists between the inside and outside surfaces of the brass vessel if the water is cooling at 1kelvin per second?
Answer:-
We know that
"Rate = {kA(\\Delta T) \\over d}" ......(1)
k=conductivity (for brass =110 w/m/oc)
"\\Delta T" = temperature difference
= T2-T1 (T2 is outside temperature and T1 is inside temperature )
d= thikness
And also we know
Q = mc"\\Delta T"
m=density ×volume
m=997kg/m3×5×10-3m3
=4.9kg
=4900g
C=4.18 joule/gram
"\\Delta T" =1k
Q= 4900×4.18×1
Q=20,482.
Putting values in (1)
We
20482="{110\u00d70.2(\\Delta T) \\over 4\u00d710^{-3}}"
"\\Delta T =3.7" answer
that is temperature difference
Comments
Leave a comment