Liquid is filled in a vessel which is kept in a room with temperature $20^{\circ}C$ . When the temperature of the liquid is $80^\circ \mathrm{C}$ , then it loses heat at the rate of 60 cal/sec . What will be the rate of loss of heat when the temperature of the liquid is $40^\circ \mathrm{C}$
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Rate of loss of heat $\left(\frac{\Delta Q}{t}\right) \propto$ temperature difference Δ Δ θ θ
$\frac{\left(\frac{\Delta Q}{t}\right)_1}{\left(\frac{\Delta Q}{t}\right)_2} = \frac{\Delta \theta_2}{\Delta \theta_1}$ ⇒ ⇒ $\frac{60}{\left(\frac{\Delta Q}{t}\right)_2} = \frac{80-60}{40-20}$ ⇒ ⇒ $\left(\frac{\Delta Q}{t}\right)_2 = 20 \frac{\mathrm{cal}}{\mathrm{sec}}$
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