Equal volumes of water and alcohol when put in similar calorimeters take `100 sec` and `74 sec` respecively to cool from `50^(@)C` to `40^(@)C`. The thermal capacity of each calorimeter in numerically equal to the volume of either liquid. The specific gravity of alcohol is `0.8` if the specific heat capacity of water is `1 cal//gm`, the specific heat capcity of alcohol will

Options

A. `0.6 cal//gm-^(@)C`
B. `0.8 cal//gm-^(@)C`
C. `1.6 cal//gm-^(@)C`
D. `1.8 cal//gm-^(@)C`

Answer

A

Explanation

Let `V` c c be the volume of either liquid.
Theremal capacity of each calorimeter is `V "per" ^(@)C`
Mass of water `V xx 1 = Vg`
Mass of alchol = `V xx 0.8 = 0.8Vg`
Rate of cooling water,
`=1/100[V(50-40)+V xx 1 xx (50-40)]=(V)/(5) cal//sec`
Rate of cooling of alcohol
`=1/74[V(50-40)+S+0.8V xx 10]`
`1/74[10V+8VxxS]`
Now, `1/74 [10V+8VxxS]=V/5`
`:. s=0.6 cal//g-^(@)C`.