Question
Rays from the sun are focused by a lens of diameter $5 \mathrm{~cm}$ on to a block of ice and $10 \mathrm{~g}$ of ice is melted in 20 min. Therefore the heat from the sun reaching the earth per min per square centimetre is (Latent heat of ice $L=80 \mathrm{cal} \mathrm{g}^{-1}$ )(a) $2.04 \mathrm{cal}$(b) $0.51 \mathrm{cal}$(c) $4.08 \mathrm{cal}$(d) $3.02 \mathrm{cal}$
Step 1
This can be represented by the equation: \[ Q = mL \] where \(m\) is the mass of the ice, \(L\) is the latent heat of fusion, and \(Q\) is the total heat. Show more…
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The rays of sun are focussed on a piece of ice through a lens of diameter $5 \mathrm{~cm}$, as a result of which $10 \mathrm{~g}$ ice melts in 10 min. The amount of heat received from sun, per unit area per min is (a) $4 \mathrm{cal} \mathrm{cm}^{-2} \min ^{-1}$ (b) $40 \mathrm{cal} \mathrm{cm}^{-2} \mathrm{~min}^{-1}$ [c) $4 \mathrm{Jm}^{-2} \mathrm{~min}$ (d) $400 \mathrm{cal} \mathrm{cm}^{-2} \mathrm{~min}^{-1}$
Heat and Kinetic Theory of Gases
Round 1
It takes 20 minutes to melt $10 \mathrm{~g}$ of ice, when rays from the 'sun are focussed by a lens of diameter $5 \mathrm{~cm}$ on to a block of ice. The heat received from the sun on $1 \mathrm{~cm}^{2}$ per minute is : (Given: $L=80 \mathrm{k} \mathrm{cal} / \mathrm{kg}$ ) (a) $R=2.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}$ (b) $R=3.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}$ (c) $R=0.204 \mathrm{cal} / \mathrm{cm}^{2}-\min$ (d) $R=204 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}$
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