Thermal Properties of Matter

Step: Find the specific heat.
 \(Q= ms \Delta T~~~…(1)\)
\(i=\dfrac{dQ}{dt}=\text{constant}\)
now, from equation \((1)\) we will get
\(\dfrac{dQ}{dt}=ms\left(\dfrac{d T}{dt}\right)\)
\(\dfrac{i}{ms}=\dfrac{dT}{dt}\)
\(\Rightarrow\dfrac{d T}{dt} = \text{slope of the graph}\propto \dfrac{1}{ms}\)
Therefore, the more of the specific heat, the lesser the slope; the lesser of the specific heat, the more of the slope.
Hence, option (2) is the correct answer.