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.

Question

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.

Infinity Learn · Accepted Answer

$$m1=mass$$ 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 $$liquidm1S1$$Δθ$$1=m2S2$$Δθ$$2+m3S3$$Δθ$$31000$$×$$12$$×$$(80-40)=200$$×$$12(40-20)+900$$×$$S3(40-20)Solving$$ this equation we get, $$S3=1cal/g$$·C  ∘

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