Question

A steel vessel with a bottom thickness $3.2\text{cm}$ is placed on a hot stove. The area of the bottom of the vessel is $0.32{\text{m}}^2$. The water inside the vessel is at ${100}^{\circ }C$ and $0.5\text{kg}$ are evaporated every $4$ minute. Find the temperature of the lower surface of the vessel, which is in contact with the stove.
(Take latent heat of vapourization, $L_{\text{v}}=2.4\times {10}^6\text{J/kg}$ and thermal conductivity of steel, $k_{\text{steel}}=50\text{W/mK}$)

• A. ${100}^{\circ }C$
• B. ${110}^{\circ }C$
• C. ${120}^{\circ }C$
• D. ${130}^{\circ }C$

Answer 1

The correct option is B ${110}^{\circ }C$

Given that,
$L_{\text{v}}=2.4\times {10}^6\text{J/kg}$

$k_{\text{steel}}=50\text{W/mK}$

$T_1={100}^{\circ }C$

Thickness, $L=3.2\text{cm}=3.2\times {10}^{-2}\text{m}$
Area, $A=0.32{\text{m}}^2$

Rate of heat received by water,

$\frac{d{Q}_1}{dt}=L_{\text{v}}\left(\frac{dm}{dt}\right)$

$\Rightarrow \frac{d{Q}_1}{dt}=2.4\times {10}^6\times \frac{0.5}{4\times 60}=5\times {10}^3\text{J/s}$

Rate of heat transfer through steel,

$\frac{d{Q}_2}{dt}=\frac{(T_2-T_1)}{L/kA}=\frac{T_2-100}{\left(\frac{3.2\times {10}^{-2}}{0.32\times 50}\right)}$

$\Rightarrow \frac{d{Q}_2}{dt}=(T_2-100)\times 500$

$\because \frac{d{Q}_1}{dt}=\frac{d{Q}_2}{dt}$

$\Rightarrow 5\times {10}^3=(T_2-100)500$

$\Rightarrow T_2=110{}^{\circ }C$

Tips: Heat transfer through the steel is totally utilized by the water. Here water is at ${100}^{\circ }C$, so heat absorbed by water is consumed to evaporate it.