A 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 $20^{o} C$ . The mixture attains and equilibrium temperature of $60^{o} C$ . Find the temperature of the furnace. Specific heat capacity of iron $= 470 J$ ${k g}^{- 1}$ ${^{o} C}^{- 1}$ .

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Solution

Mass of iron = 100 g
Water equivalent of calorimeter = 10 g
Mass of water = 240 g

Let the temperature of the surface be $T^{o} C$
Specific heat of iron = S = $470 J k g^{- 1} C^{- 1}$

Total heat gained = Total heat lost
$\frac{100}{1000} \times 470 \times (T - 60) = \frac{250}{1000} (4200) (60 - 20)$
$T = {953.61}^{0} C$

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