The 'Kangri' is an earthen pot used to stay warm in Kashmir during the winter months. Assume that the 'Kangri' is spherical and of surface area $7 \times 10^{-2} \,m ^{2}$. It contains $300 g$ of a mixture of coal, wood and leaves with calorific value of $30 \,kJ / g$ (and provides heat with $10 \%$ efficiency). The surface temperature of the 'Kangri' is $60^{\circ} C$ and the room temperature is $0^{\circ} C$. Then, a reasonable estimate for the duration $t$ (in h) that the 'Kangri' heat will last is (take the 'Kangri' to be a black body)
KVPY 2020, Advanced
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$(b)$ Given, $A=7 \times 10^{-2} \,m ^{2}$
$m=300 \,g$
Calorific value $=30 \,kJ / g$
Efficiency, $\eta=10 \%=0.1$
$T_{1}=0^{\circ} C =273 \,K$
$T_{2}=60^{\circ} C =333 \,K$
$e=1$ (black body)
From Newton's law of cooling,
$\frac{d Q}{d t}=e A \sigma\left(T_{2}^{4}-T_{1}^{4}\right)$
$=1 \times 7 \times 10^{2} \times 5.735 \times 10^{8}$
$=26.75 \,W$
$\text { Total heat produced, }$
$H =\eta m \times \text { calorific value }$
$\left.=0.1 \times 300 \times 30 \times 10^{3}-(273)^{4}\right]$
$=9 \times 10^{5} \,J$
$\therefore \text { Time, }$ $t =\frac{\text { Heat produced }(H)}{\text { Rate of emission }\left(\frac{d Q}{d t}\right)}$
$=\frac{9 \times 10^{5}}{26.75 \times 3600}=9.35 \,h \simeq 10 \,h$
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