Physics - Heat & Thermo Question with Solution | TestHub

$\text{q }=\text{ mCT}$
$\frac{\text{dq}}{\text{dt}}=\text{mc}\frac{\text{dT}}{\text{dt}}$
R = rate of absorption of heat$=\frac{\text{dq}}{\text{dt}}\propto \text{C}$
(i) in$0-100 \mathrm{K}$
The graph between ($0-100 \mathrm{K}$) appears to be a straight line upto a reasonable approximation.
(ii)$\Delta \text{q}=\text{mC}\Delta \text{T}$as C  is more in ($400-500 \mathrm{K}$) then ($0-100 \mathrm{K}$) so heat is increasing.
(iii) C remains constant so there no change in R  from ($400-500 \mathrm{K}$)
(iv) C is increases so R is increases in range ($200-300 \mathrm{K}$)