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A liquid of mass m and specific heat S is heated to a temperature T. Another liquid of mass m/2 and specific heat 2S is heated to a temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture
  • a)
    2/3 T
  • b)
    8/5 T
  • c)
    3/3 T
  • d)
    3/2 T
Correct answer is option 'D'. Can you explain this answer?
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A liquid of mass m and specific heat S is heated to a temperature T. A...
Given:
- Mass of first liquid = m
- Specific heat of first liquid = S
- Temperature of first liquid = T
- Mass of second liquid = m/2
- Specific heat of second liquid = 2S
- Temperature of second liquid = 2T

To find: Resultant temperature of mixture.

Solution:
Let the resultant temperature of the mixture be x.

Heat gained by first liquid = m × S × (x - T)
Heat gained by second liquid = (m/2) × 2S × (x - 2T) = m × S × (x - 2T)

As per the law of conservation of energy, the heat gained by the two liquids must be equal to the heat lost by the mixture.

Heat lost by mixture = Heat gained by first liquid + Heat gained by second liquid
= mS(x - T) + mS(x - 2T)
= 2mS(x - T)

Equating the two, we get:
2mS(x - T) = m × S × (x - 2T) + mS(x - T)
Simplifying, we get:
2x - 2T = x - 2T + x - T
Solving for x, we get:
x = (3T + 2T)/2 = 5T/2

Therefore, the resultant temperature of the mixture is 5T/2 or 2.5T, which is option D.
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A liquid of mass m and specific heat S is heated to a temperature T. A...
Use heat loss =heat gained by body
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A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solulte is present , there is additional contribution to the entropy of the liquid due to increase randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus , a solute (non volatil e) lowers the vapour pressure of a liquid, and hence a higher booing point of the solution Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution . Elevation of B.Pt. Tband depression of F.Pt. Tfof a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identity.For dilute solutions, Tband Tfare proportional to the molality of the solute in the solution.The vaues of Kb and Kf do depend on the properties of the solvent. For liquids, is almost constant . [Troutans Rule , this constant for most of the Unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state) is equal to 90J/mol. ] For solutes undergoing change of molecular state is solution (ionization or association), the observed T values differ from the calculate ones using the above relations. In such situations, the relationships are modified as Where i = Vant Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.Q.A mixture of two immiscible liquids at a constant pressure of 1 atm boils at a temperature

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A liquid of mass m and specific heat S is heated to a temperature T. Another liquid of mass m/2 and specific heat 2S is heated to a temperature 2T. If these two liquids are mixed, the resultant temperature of the mixturea)2/3 Tb)8/5 Tc)3/3 Td)3/2 TCorrect answer is option 'D'. Can you explain this answer?
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