Question

steam at 100^o c is passed into a copper cylinder 10mm thick and of 200 cm^2 area .water at 100^o collects at the rate of 150g min ^-1 .find the temperature of the outer surface ,if the conductivity of copper is 0.8 cal s^-1 ^0c6-1 and the latent heat of steam is 540 cal g ^-1?

Answer 1

Thermal conductivity = K = 0.8 cal  s⁻¹ ⁰K⁻¹ cm⁻¹
d = thickness between two surfaces at temperatures T1 & T2 = 1 cm
ΔT = T1 - T2
A = curved surface area of cylinder
T1 = 100°C (inner surface).   T2 = ?

The heat conducted is equal to the latent heat released by condensation of steam.

Heat conducted through the cylinder (through its CSA) 
  = Q = K  ΔT * A / d
 Q =150 /60 gms/sec * 540 cal /gm = 1,350 cal/sec
 
 1,350 = 0.8 * ΔT * 200 / 1
ΔT = 8.4375°C

T2 = 100 - 8.4375 = 91.5625°C