A piece of iron of mass 100g is kept inside a furnace for a long time

1.	A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.
Answer» mass of Iron = 100g water Eq of caloriemeter = 10g mass of water = 240g Let the Temp. of surface = 0^ºC S_iron = 470J/kg°C Total heat gained = Total heat lost. So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20) => 47θ – 47 × 60 = 25 × 42 × 40 => θ = 4200 + 2820/47= 44820/47 = 953.61°C

A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.

Answer»

mass of Iron = 100g

water Eq of caloriemeter = 10g

mass of water = 240g

Let the Temp. of surface = 0^ºC

S_iron = 470J/kg°C

Total heat gained = Total heat lost.

So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20)
=> 47θ – 47 × 60 = 25 × 42 × 40
=> θ = 4200 + 2820/47= 44820/47 = 953.61°C

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A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.

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