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A liquid initially at 30 deg C is added very slowly into a calorimeter whose temperature is 110 deg C. The boiling point of the liquid is 80 deg C. The first 5 g of liquid poured in evaporates completely. After a further 80 g of the same liquid is added, the system settles at an equilibrium temperature of 50 deg C. Assuming no heat is lost to the surroundings, what is the ratio of the latent heat of the liquid to its specific heat (in deg C)?

1. 180 deg C
2. 270 deg C
3. 200 deg C
4. 240 deg C

Correct answer: 270 deg C

Solution

Let the calorimeter's total heat capacity be W (in cal/deg C, but it cancels) and s = specific heat of liquid, L = latent heat. Stage 1: calorimeter cools 110->80, and the 5 g first warms 30->80 then fully vaporises. Stage 2: calorimeter (plus the residue) cools 80->50 while 80 g liquid warms 30->50. Solving the two equations gives L/s = 270 deg C.

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