Two substances A and B of equal mass are heat | vedclass

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(C) The heat absorbed during the phase change (melting) is given by $H = mL$,where $m$ is the mass and $L$ is the latent heat of fusion.Since the subs

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$8/5$

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(C) The heat absorbed during the phase change (melting) is given by $H = mL$,where $m$ is the mass and $L$ is the latent heat of fusion.Since the substances are heated at a constant rate $P = 6 \ cal \ s^{-1}$,the heat absorbed is $H = P 	imes \Delta t$,where $\Delta t$ is the time interval during which the substance is melting (the horizontal part of the graph).For substance $A$,the melting occurs from $t = 2 \ s$ to $t = 6 \ s$,so $\Delta t_A = 6 - 2 = 4 \ s$.For substance $B$,the melting occurs from $t = 4 \ s$ to $t = 6.5 \ s$,so $\Delta t_B = 6.5 - 4 = 2.5 \ s$.Since the masses are equal,the ratio of heat absorbed is $H_A/H_B = (P 	imes \Delta t_A) / (P 	imes \Delta t_B) = \Delta t_A / \Delta t_B$.$H_A/H_B = 4 / 2.5 = 40 / 25 = 8/5$.

Column-$I$	Column-$II$
$(a)$ Required heat to convert solid into gaseous.	$(i)$ Latent heat of fusion
$(b)$ Required heat to convert solid into liquid.	$(ii)$ Latent heat of vaporization

Two substances $A$ and $B$ of equal mass are heated at a constant rate of $6 \ cal \ s^{-1}$. The graph shows the temperature versus time. What is the ratio of the heat absorbed for complete melting $H_A/H_B$?

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