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Calorimetry

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By Expert Faculty of Sri Chaitanya
Question

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cc of water at 27°C. The final temperature is 40°C. If heat losses to the surroundings is not negligible, then the value of specific heat (in cal/g°C) of the metal will be_____

Moderate
Solution

Here, mass of metal, m = 0.20 kg =200 g
Fall in temperature of metal

           ΔT=150C40C=110C

Ife is specific heat of the metal, then heat lost by the metal,

                     ΔQ=mcΔT=200s×110   ……….(i)

Volume of water = 150 cc
Mass of water, m' = 150g
Water equivalent of calorimeter

                         w=0.025kg=25g

Rise in temperature of water in calorimeter

ΔT=40C27C=13C

Heat gained by water and calorimeter

      ΔQ=m+wΔT=(150+25)×13      ΔQ=175×13    .........(ii)As ΔQ=ΔQ

 From Eqs. (i) and (i), we get

200×s×110=175×13s=175×13200×1100.10 J/gC


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The mass, specific heat capacity and the temperature of a solid are 1000 g12  cal g.C  -1 and 80°C respectively. The mass of the liquid and the calorimeter are 900 g and 200 g. Initially, both are at room temperature 20°C. Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is 40°C, then find the specific heat capacity of the unknown liquid.

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