8.0 g of cream at 17.0 °C are added to an insulated cup containing 150.0 g of coffee at 73.2 °C. Calculate the equilibrium temperature of the coffee.

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

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

there are two significant figures.