Question

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 1

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=${110×0.2(\Delta T) \over 4×10^{-3}}$

$\Delta T =3.7$ answer

that is temperature difference