Question

A steel ball (radius 25mm) of uniform temperature 700k is immersed in water whose temperature is 400k. calculate the temperature of ball after 1 hr. given k =40w/mk. , h=11 w/m2k, ρball = 7850 kg/m3 & cp=0.45kj/kgk.

Answer 1

Given, radius of steel ball , r = 25mm
so, volume of steel ball , V = 4/3πr³
= 4/3 π ( 0.025)²
and area of steel ball, A = 4π(0.025)²

T = 400K , Ti = 700K , t = 1 hr , h = 11w/m²k
k = 40 w/mK , ρ = 7850 kg/m³ and Cp = 0.45kJ/kg.K

now, use formula, $\frac{T-T_{co}}{T_i-T_{co}}=e^{-(\frac{hA}{\rho C_pV})t}$

so, ln{(Ti - Tco)/(T - Tco)} = (hA/ρCpV)t

=> ln(700 - Tco)/(400 -Tco) = {11 × 4π(0.025)²/{7850 × 0.45 × 4/3π(0.025)³} × 1

=> Tco =-262.16k