Question

Water is being boiled in a flat bottomed kettle placed on a stove. The area of the bottom is $300cm$${}^2$ and the thicknes is $2mm$. If the amount of steam produced is $1gm$ min${}^{-1}$, then the difference of the temperature between the inner and the outer surface of the bottom is(thermal conductivity of the material of the kettle $0.5calcm$${}^{-1}$s${}^{-1}$C${}^{-1}$, latent heat of the steam is equal to $540calg$${}^{-1}$)

• A. ${12}^{o}C$
• B. ${1.2}^{o}C$
• C. ${0.12}^{o}C$
• D. ${0.012}^{o}C$

Answer 1

Answer: (D)

${0.012}^{o}C$

Let the temperature difference be $t$,

so the rate of heat conduction is given as $Q/t=KAt/l$ where $K=0.5cal$ is the thermal conductivity, $A=300c{m}^2$ is the area of surface, $l=0.2cm$ is the thickness.

Now rate of vaporization is $m_l=1gmmi{n}^{-1}$, so heat absorption rate is $1\times 540/60=9cal{s}^{-1}$

Equating the heat rate, we get $0.5\times 300\times t/0.2=9\Rightarrow t={0.012}^{o}C$