Question

The rate of cooling of 100g of water at ${77}^{o}C$ 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
• B. 80 cal/min
• C. 89 cal/min
• D. none of the above

Answer 1

The correct option is C 89 cal/min

In this question formula used,

Rate of cooling=$\frac{dT}{dt}=\frac{eA\sigma }{mc}(T^4-{T_0}^4)$

Where,(e=emissivity, A=Area, $\sigma$=Stefan's constant, m=mass, C=specific heat)

So for water

$\frac{-dT}{dt}=\frac{eA\sigma }{mc}[T^4-{T_0}^4]$ (T+Temperature of body, $T_0$=Temperature of surrounding)

$80=\frac{eA\sigma }{mc}[T^4-{T_0}^4]$ (1)

So for turpentine

$\theta =\frac{eA\sigma }{mTc}[T^4-{T_0}^4]$ (2)

(e=same for both due to same calorimeter. C=same for both due to same calorimeter)

Divide 1 and 2

$\frac{80}{\theta }=\frac{mT}{mw}=\frac{dT\times vT}{dw\times vw}[vT=vw]$

$\frac{80}{\theta }=0.9$

$\frac{800}{9}=\theta$

Rate of cooling of turpentine oil =89cal/minute