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# The mass, specific heat capacity and the temperature of a solid are , $\frac{1}{2}$   and $80°\mathrm{C}$ respectively. The mass of the liquid and the calorimeter are  and . Initially, both are at room temperature $20°\mathrm{C}$. Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is $40°\mathrm{C}$, 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  ∘
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detailed solution

Correct option is C

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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In electrical calorimeter experiment, voltage across the heater is  and current is . Heater is switched on for . Room temperature is ${\theta }_{0}=10.0°\mathrm{C}$ and final temperature of calorimeter and unknown liquid is ${\theta }_{f}=73.0°\mathrm{C}$. Mass of empty calorimeter is  and combined mass of calorimeter and unknown liquid is . Find the specific heat capacity of the unknown liquid in proper significant figures. Specific heat of calorimeter =

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