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 $20^{\circ} C$ . The mixture attains an equilibrium teperature of $60^{\circ} C .$ Find the temperature of the furnace. Specific heat capacity of iron $= 470 H k g^{- 1}^{\circ} C^{- 1} .$

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Solution

Given, Mass of iron = 100g

Water equivalent of calory meter

= 10 g

Mass of water = 240 gm

Let the temp of surface $= 0^{\circ} C$

$S_{i r o n} = 470 - J / k g$

Total heat gained = Total heat lost

So, $\frac{100}{1000} \times 470 \times (θ - 60^{\circ})$

$= \frac{(240 + 10)}{1000} \times 4200 \times (60 - 20)$

$\Rightarrow 47 θ - 47 \times 60 = 25 \times 42 \times 40$

$\Rightarrow θ = \frac{42000 + 2820}{47}$

$= \frac{44820}{47} = {953.61}^{\circ} C$

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