Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains 20 kg of water. A 2.3 kg piece of the metal initially at a temperature of 165°C is dropped into the water. The container and water initially have a temperature of 15.0°C, and the final temperature of the entire system is 18.0°C.

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg) (C) (147°C) = (3.8 kg) (C) (3°C) + (20 kg) (4186 J/kg°C) (3°C)

C[ (2.3 kg) (147°C) - (3.8 kg) (3°C) ] = 252000 J

C = 252000 J/326.7 kg°C

C = 771.35 J/kg°C