80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.​

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

Therefore, the resulting temperature is 23.37 ⁰C

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

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