Q.
The mass, specific heat capacity and the temperature of a solid are 1000 g, 12 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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a
10 cal/g·C ∘
b
100 cal/g·C ∘
c
1 cal/g·C ∘
d
0.1 cal/g·C ∘
answer is C.
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Detailed Solution
m1=mass of solid=1000 gS1=specific heat of solid=12cal/g-C ∘=S2 or specific heat of calorimeterm2= mass of calorimeter =200 gm3= mass of unknown liquid = 900 gS3= specific heat of unknown liquidFrom law of heat exchange,Heat given by solid = Heat taken by calorimeter + Heat taken by unknown liquidm1S1Δθ1=m2S2Δθ2+m3S3Δθ31000×12×(80-40)=200×12(40-20)+900×S3(40-20)Solving this equation we get, S3=1cal/g·C ∘
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