Question

An unknown substance has a mass of 0.125 kg and an initial temperature of 90.5°C. The substance is then dropped into a calorimeter made of aluminum containing 0.285 kg of water initially at 29.5°C. The mass of the aluminum container is 0.150 kg, and the temperature of the calorimeter increases to a final equilibrium temperature of 32.0°C. Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.

Answer 1

Answer:

The specific heat capacity of the substance = 455.38 J/kgK

Explanation:

Heat lost by the substance = Heat gained by water + heat gained by the aluminum calorimeter

Qs = Qw + Qc.................... equation 1

Where Qs = heat lost by the substance, Qw = heat gain by water, Qc = heat gain by the aluminum calorimeter.

Qs = c₁m₁(T₁-T₃)................ equation 2

Qw = c₂m₂(T₃-T₂)............. equation 3

Qc = c₃m₃(T₃-T₂)............. equation 4

Where c₁ = specific heat capacity of the substance, m₁ = mass of the substance, c₂ = specific  heat capacity of water, m₂ = mass of water, c₃ = specific heat capacity of aluminium, m₃ = mass of the aluminum container, T₁ = Initial Temperature of the substance, T₂ = initial temperature of water, T₃ = Final equilibrium temperature.

Substituting equation 2, 3, 4 into equation 1

c₁m₁(T₁-T₃) = c₂m₂(T₃-T₂) + c₃m₃(T₃-T₂)................. equation 5

Making c₁ the subject of equation 5

c₁ =  {c₂m₂(T₃-T₂) + c₃m₃(T₃-T₂)}/m₁(T₁-T₃)............... equation 6

Where c₂ = 4200 J/kgK, m₂ = 0.285 kg, m₁ = 0.125 kg, c₃ = 900 J/kgK, m₃= 0.150 kg, T₁ = 90.5°C, T₂ = 29.5°C, T₃ =  32.0°C

Substituting these values into Equation 6,

c₁ = {4200×0.285(32-29.5) + 900×0.150(32-29.5)}/0.125(90.5-32)

c₁ = {1197(2.5) + 135(2.5)}/7.3125

c₁ = {2992.5 + 337.5}/7.3125

c₁ = 3330/7.3125

c₁ = 455.38 J/kgK.

Therefore the specific heat capacity of the substance = 455.38 J/kgK