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(B) Assuming no heat loss,the heat gained by the liquid in the calorimeter is equal to the heat supplied by the heater.$Q = m S \Delta T = P t$Where $

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the heat capacity of the liquid is inversely proportional to the slope of the graph

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(B) Assuming no heat loss,the heat gained by the liquid in the calorimeter is equal to the heat supplied by the heater.$Q = m S \Delta T = P t$Where $m$ is the mass of the liquid,$S$ is the specific heat capacity,and $T_f - T_i = \Delta T$.$\Rightarrow m S (T_f - T_i) = P t$$\Rightarrow T_f = \left( \frac{P}{m S} ight) t + T_i$Comparing this with the equation of a straight line $y = mx + c$,we get:Slope $= \frac{P}{m S} = \frac{P}{C}$,where $C = m S$ is the heat capacity of the liquid.Since the slope is $\frac{P}{C}$,the heat capacity $C$ is inversely proportional to the slope of the graph $(C \propto \frac{1}{	ext{slope}})$.Therefore,option $(b)$ is correct.

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