A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20 °C. The mixture attains an equilibrium temperature of 60 °C. The temperature of the furnace is

400 ^oC

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406.80 ^oC

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506.80 ^oC

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500 ^oC

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Solution

The correct option is C
506.80 ^oC

Let θ₁ be temperature of the furnace that is temperature of the piece of iron.

Thus, the heat lost by piece of iron is

Q = M₁C₁ (θ₁ - θ)

= 0.2 x 470 (θ₁ - 60) = 94 (θ₁ - 60)

Heat gained by water and calorimeter is

Q = (M₂ + w) x c₂ x (θ - θ2)

= (0.24 + 0.01) x 4200 x (60 - 20) = 42000

From above two equations

94 (θ₁ - 60) = 42000

θ₁ = 506.80 ^oC

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