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Question

The rate of cooling of 100g of water at 77oC is 80cal/min. If the water is replaced by an equal volume of turpentine oil in the same calorimeter and allowed to cool in the same environment, then the rate of loss of heat of turpentine oil will be (Given, relative density of oil = 0.9).

A
72 cal/min
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B
80 cal/min
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C
89 cal/min
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D
none of the above
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Solution

The correct option is C 89 cal/min
In this question formula used,
Rate of cooling=dTdt=eAσmc(T4T04)
Where,(e=emissivity, A=Area, σ=Stefan's constant, m=mass, C=specific heat)
So for water
dTdt=eAσmc[T4T04] (T+Temperature of body, T0=Temperature of surrounding)
80=eAσmc[T4T04] (1)
So for turpentine
θ=eAσmTc[T4T04] (2)
(e=same for both due to same calorimeter. C=same for both due to same calorimeter)
Divide 1 and 2
80θ=mTmw=dT×vTdw×vw[vT=vw]
80θ=0.9
8009=θ
Rate of cooling of turpentine oil =89cal/minute


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