Answer:
70.35°C
Explanation:
From the question,
Assuming: The specific heat capacity of cream and coffee is thesame as that of water = 4.2 J/g.°C
Applying
Heat lost by the coffee = heat gained by the cream
cm(t₁-t₃) = cm'(t₃-t₂)................ Equation 1
Where c = specific heat capcity of both cream and coffee, m = mass of coffee, t₁ = intial temperature of coffee, m' = mass of cream, t₂ = initial temperature of cream, t₃ = equilibrium temperature
From the question,
Given: m = 150 g, t₁ = 73.2°C, m' = 8 g, t₂ = 17°C
Constant: c = 4,2 J/g.°C
Substitute these values into equation 1
4.2×150(73.2-t₃) = 4.2×8(t₃-17)
630(73.2-t₃) = 33.6(t₃-17)
Solve for t₃
46116-630t₃ = 33.6t₃-571.2
630t₃+33.6t₃ = 571.2+46116
663.6t₃ = 46687.2
t₃ = 46687.2/663.6
t₃ = 70.35°C
