Question

A small copper ball of $5mm$ diameter at $500K$ is dropped into an oil bath whose temperature is $300K.$ The thermal conductivity of copper is $400W/mK,$ its density $9000kg/m^3$ and its specific heat $385J/kgK.$ If the heat transfer coefficient is $250W/m^{2}K$ and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be , in $K/s,$

• A.
  $8.7$
• B.
  $13.9$
• C.
  $17.3$
• D.
  $27.7$

Answer 1

The correct option is C

$17.3$
Given data:

$k=400W/mK;T_i=500K;\rho =9000kg/m^3{T}_{\infty }=300K;C_p=985J/kgK:h=250W/m^{2}K$

$L_c=\frac{V}{A}=\frac{D}{6}=\frac{0.005}{6}$
$=8.33\times {10}^{-4}m$

$T=T_{\infty }+(T_i-T_{\infty })e^{\frac{-hAt}{\rho cV}}$

$\frac{dT}{dt}=\frac{-hA}{\rho cV}(T_i-T_{\infty )}{e}^{\frac{-hAt}{\rho cV}}$


${\begin{array}{c}\frac{dT}{dt}\end{array}|}_{t=0}=\frac{hA}{\rho cV}(T_i-T_{\infty )}[magnitude]$
$=\frac{250\times 200}{9000\times 385\times 8.33\times {10}^{-4}}$

${\begin{array}{c}\frac{dT}{dt}\end{array}|}_{t=0}=17.31K/s$