Questions

An insulated container, there are two fluid with specific heat capacities c_{1} and c_{2} , separated by insulating barrier. Liquid temperatures are different. The partition is removed and after the establishment of thermal equilibrium, the difference between initial temperature of first liquid and temperature established in the vessel is two times smaller than the difference between the initial temperatures of liquids. Find the mass ratio of the first and second fluid.

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a

c2c1

b

c1c2

c

c1c2+1

d

2c2c1+1

NEW

detailed solution

Correct option is A

T1−T0=T1−T22⇒T0=T1+T22 Also, m1c1T1−T0=m2c2T0−T2⇒m1c1T1−T22=m2c2T1−T22⇒m1m2=c2c1

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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