Question

Water is boiled in a container having a bottom of surface are $25c{m}^2$, thickness 1.0 mm and thermal conductivity 50 $W{m}^{-1\circ }{C}^{-1}$. 100 g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = $2.26\times {10}^{6}JK{g}^{-1}$.

Answer 1

A = 25 $c{m}^2=25\times {10}^{-4}{m}^2$,

L = 1 mm =${10}^{-3}m,K=50W/m^{\circ }C$

$\frac{Q}{t}$= rate of conversion of water in to steam

$\frac{Q}{t}=\frac{100\times {10}^{-3}\times 2.26\times {10}^6}{1min.}$

$=\frac{100\times {10}^3\times 2.26}{1min}$

$=\frac{2.26}{6}\times {10}^4$

$=0.376\times {10}^{4}J/s$

$\frac{Q}{t}=\frac{kA(T_1-T_2)}{l}$

$\Rightarrow 0.376\times {10}^4=\frac{50\times 25\times {10}^{-4}\times (T-100)}{{10}^{-3}}$

$\Rightarrow (T-100)=\frac{{10}^{-3}\times 0.376\times {10}^4}{50\times 25\times {10}^{-4}}$

= $\frac{{10}^{-3}\times 0.376}{50\times 25}$

$\Rightarrow (T-100)=3.008\times {10}^4\times {10}^{-3}\times {10}^8$

$\Rightarrow (T-100)=3.008\times 10$

=${30}^{\circ }C=30$

$\therefore T=100+30={130}^{\circ }$.