Two bodies of equal masses are heated at a uniform rate under identical conditions. The change in temperature in the two cases in shown graphically. What are their melting points?
Find the ratio of their specific heats and latent heats.

Explanation

(i) `A_(1)B_(1)` and `A_(2)B_(2)` represent change of state for substances 1 and 2. During change of state, temperature remains same.
Melting point of substance `1:40^(@)`C
Melting point of substance `2:60^(@)`C
(ii) Heat `DeltaQ alpha`time
Since hat is supp,ied at constant rate
`DeltaQ`=kt, k: constant rate (J/s)
`DeltaQ`=mL
mL=kt `rArr``L alpha t`
`L_(1)/L_(2)=t_(1)/t_(2)`=(time during change of state from `A_(1)` to `B_(1)`)/(time during `A_(2)` to `B_(2)`)
=(7-4)/(6-2) = 3/4
The ratio of latent heat of fusion of two substances in the ratio `3:4`
(iii) `DeltaQ=mcDeltaT rArr` kt=mc(T-`T_(0)`)
T=(kt)/(mc) +`T_(0)`, T versus t is straight line
So, k/(mc) represents slope of inclined line
In solid state, `(k/(mc))_(1)/(k/(mc))_(2) = ("slope of"B_(1)C_(1))/("slope of"B_(2)C_(2))`
`=(40//4)/(60//2) = 1/3`
`therefore` Ratio of specific heat `(c_(1)/c_(2))=1/3`
In liquid state,
`(k/(mc))_(1)/(k/(mc))_(2)` = (slope of `B_(1)C_(1)`)/(slope of `B_(2)C_(2))`
`=((70-40)//(10-7))/((100-60)//(8-6))=1/2`
Ratio of specific heat `c_(2)/c_(1)`=1/2