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

definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.