NEET-XII-Physics
25: Calorimetry
- #2A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J kg-1 °C-1.Ans : Given:
Mass of iron = 100 g
Water equivalent of calorimeter = 10 g
Mass of water = 240 gm
Let the temperature of surface be `` \theta `` °C.
Specific heat capacity of iron = 470 J kg-1 °C-1
Total heat gained = Total heat lost
`` \Rightarrow \frac{100}{1000}\times 470\times \left(\theta -60°\right)=\frac{\left(240+10\right)}{1000}\times 4200\times \left(60-20\right)``
`` \Rightarrow 47\theta -47\times 60=25\times 42\times 40``
`` \Rightarrow \theta =\frac{42000+2820}{47}=\frac{44820}{47}``
`` =953.61°\,\mathrm{\,C\,}``
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