Answer:
1205.77 J/kg.K
Explanation:
Heat lost by alloy = heat gained by water + heat gained by the calorimeter
c₁m₁(t₂-t₃) = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)................. Equation 1
Where c₁ = specific heat capacity of the alloy, m₁ = mass of the alloy, t₂ = initial temperature of the alloy, t₃ = equilibrium temperature, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of water and calorimter, c₃ = specific heat capacity of calorimter, m₃ = mass of calorimter.
Making c₁ the subject of the equation,
c₁ = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)/m₁(t₂-t₃)........................ Equation 2
Given: c₂ = 4190 J/kgK, m₂ = 250 g = 0.25 kg, m₁ = 75 g = 0.075 kg, m₃ = 55 g = 0.055 kg, c₃ = 377 J/kg.K, t₁ = 18 °C, t₂ = 100 °C, t₃ = 24.4 °C.
Substitute into equation 2
c₁ = [0.25×4190×(24.4-18) + 0.055×377×(24.4-18)]/[0.075(100-24.4)]
c₁ = (6704+132.704)/5.67
c₁ = 6836.704/5.67
c₁ = 1205.77 J/kg.K
Thus the specific heat capacity of the alloy = 1205.77 J/kg.K
