Question

A piece of iron of mass 100 g is kept inside a furnace for a long time and jthen put in a calorimeter of water equivalent 10 g containing 240 g 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 = 470 j/kg-°c.

Answer 1

mass of Iron = 100g

water Eq of caloriemeter = 10g

mass of water = 240g

Let the Temp. of surface = 0ºC

Siron = 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

Answer 2

Answer:

953.61 degrees celsius

Explanation:

mass of Iron = 100g

water Eq of calorimeter = 10g

mass of water = 240g

Let the Temp. of surface = 0ºC

Siron = 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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