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Question

Equal volumes of water and alcohol when put in similar calorimeters take 100 sec and 74 sec respectively, to cool from 50C to 40C. The thermal capacity of each calorimeter is 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/gC, the specific heat capacity of alcohol will be -
[Assume constant rate of cooling]

A
0.6 cal/gC
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B
0.8 cal/gC
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C
1.6 cal/gC
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D
1.8 cal/gC
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Solution

The correct option is A 0.6 cal/gC
Let V (in cc) be the volume of either liquid.
Then, thermal capacity of each calorimeter is V cal/C
Given,
Mass of water mw=V×ρw=V×1=V grams
Mass of alcohol ma=V×ρa=V×0.8=0.8V grams
Time taken by water to cool from 50C to 40C is (Δt1)=100 sec
Time taken by alcohol to cool from 50C to 40C is (Δt2)=74 sec
We know that,
Rate of cooling of water (dQw)dt=Heat released by waterTime taken+Heat released by calorimeter-1Time taken
=(mwcw×ΔTwΔt1)+(mvcv×ΔTvΔt1)
(dQw)dt=1100[V×1×(5040)+V×(5040)]
[given mw=V,mvcv=V]
=V5cal/sec

Rate of cooling of alcohol (dQa)dt=Heat released by alcoholTime taken+Heat released by calorimeter-2Time taken
=(maca×ΔTaΔt2)+(mvcv×ΔTvΔt2)
(dQa)dt=174[s×0.8V×(5040)+V×(5040)]
[given ma=0.8V,mvcv=V,s=specific heat capacity of alcohol]
=174[10V+8V×s] cal/sec

Given, Rate of cooling of alcohol = Rate of cooling of water
174[10V+8V×s]=V5
s=0.6 cal/gC
Thus, option (a) is the correct answer.

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