The figure given below shows the cooling curv | vedclass

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(D) Let $P$ be the rate of heat loss (power) per unit time.During the cooling phase from $A$ to $B$,the temperature drops from $90^{\circ}C$ to $80^{\

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$20$

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(D) Let $P$ be the rate of heat loss (power) per unit time.During the cooling phase from $A$ to $B$,the temperature drops from $90^{\circ}C$ to $80^{\circ}C$ in $t_1 = 2 	ext{ min}$. The heat lost is given by $Q_1 = mC\Delta T = mC(90 - 80) = 10mC$.Since $Q_1 = P 	imes t_1$,we have $10mC = P 	imes 2$,which implies $P = 5mC$.During the solidification phase from $B$ to $D$,the temperature remains constant at $80^{\circ}C$ for $t_2 = 4 	ext{ min}$. The heat lost is the latent heat,given by $Q_2 = mL$.Since $Q_2 = P 	imes t_2$,we have $mL = P 	imes 4$.Substituting the value of $P$ from the first equation into the second:$mL = (5mC) 	imes 4$$mL = 20mC$$\frac{L}{C} = 20$.

Column-$I$	Column-$II$
$A$. Conversion of liquid into solid is	$1$. Regelation
$B$. Conversion of liquid into vapour is	$2$. Sublimation
$C$. Conversion of solid into vapour directly is	$3$. Fusion
$D$. Melting of ice caused by pressure is	$4$. Vaporisation

The figure given below shows the cooling curve of a pure wax material after heating. It cools from $A$ to $B$ and solidifies along $BD$. If $L$ and $C$ are the respective values of latent heat and the specific heat of the liquid wax,the ratio $L/C$ is

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