Question

A sphere of 0.047 kg aluminum is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transferred to a0.14 kg copper calorimeter containing 0.25 kg water at 20°C. The temperature of water rises and attains a steady state at 23°C. Find specific heat capacity of aluminum.

Answer 1

To Find↷

• specific heat capacity of aluminum

Given↷

• Mass of Al sphere (m₁)= 0.047 kg
• Initial temperature=100°C
• Final temperature = 20°C
• ∆T₁= (100°-23°) C= 77°C
• Mass of H₂O (m₂)= 0.25 kg
• Mass of Calorimeter(m₃)=0.14 Kg
• Initial temperature for calorimeter& H₂O = 20°C
• Final temperature for calorimeter& H₂O = 23°C
• ∆T₂ = 23°C-20°C = 3°C

Solution ↷

We know that,

Heat lost by the sphere = Heat gained by H₂O & Calorimeter

=> m₁ x c₁ x ∆T₁ = m₂ x c₂ x ∆T₂ + c₃ x m₃ x ∆T₂

=> 0.047 kg x c₁ x 77°C = 0.25 kg x 4.18 x 3°C+0.14 x 0.386 x 3°C

=> c₁ = ( 0.25 kg x 4.18 kg^-1 k^-1 x 3°C + 0.14kg x 0.386 kg^-1 k^-1 x 3°C ) / 0.047 kg x 77°C

=> c₁ = 3.30/ 0.047 x 77

=> c₁ = 0.913 KJ kg^-1 k^-1

Hence , the specific heat capacity of aluminum would be 0.913 KJ kg^-1 k^-1