Question

A container having volume $282.7c{m}^3$ and total surface area $245c{m}^2$ is completely filled with milk whose initial temperature is ${25}^{\circ }C.$ The continually stirred milk container is suddenly expose to steam bath at ${100}^{\circ }C.$ The overall heat transfer coefficient between steam and milk is $1136W/m^{2}K.$
The properties of milk are:
Specific heat capacity $=3.9kJ/kgK$
Thermal conductivity $=0.54W/m-K$
Density $=1030kg/m^3.$
Neglecting thermal resistance and heat capacity of container walls the necessary time required in seconds to heat milk up to the temperature of ${85}^{\circ }C,$ will be ____________

• A. 65.66 sec
• B. 60 sec
• C. 656.6 sec
• D. 75 sec

Answer 1

The correct option is A 65.66 sec

Lumped heat capacity analysis can, be applied because thermal resistance of container walls are neglected.

$\text{Given:}{T}_i={25}^{\circ }C,T_{\infty }={100}^{\circ }C,T={85}^{\circ }C,h=1136W/m^{2}K,V=202.7c{m}^3,A=245c{m}^2,C_p=3.9kJ/kg.K,\rho =1030kg/m^3\frac{T_i-T_{\infty }}{T-T_{\infty }}=e^{(\frac{hA}{\rho V{c}_p}\times \tau )}\frac{25-100}{85-100}=e^{(\frac{1136\times 245\times {10}^{-4}}{1030\times 282.7\times {10}^{-6}\times 3900}\times \tau )}5=e^{\left(0.02451025\tau \right)}\text{Taking log on both sides}ln5=0.0245\tau \Rightarrow \tau =65.66second$