What would be the appropriate equilibrium temperature of a system obtained by adding 25.0 g of ice at 0 °C to 250.0 mL of "hot" (80 °C)coffee,assuming that the heat capacity and density of the coffee are the same as for pure water (4.18 J g^-1 °C^-1 and 0.997 g mL^-1),and also assuming negligible heat transfer with the surroundings? [For water,Δ_fusH = 6.02 kJ/mol.] A)33 °C B)40 °C C)65 °C D)73 °C E)79 °C

Question

Quizplus · Accepted Answer

First, we need to calculate the heat required to melt the ice:
q1 = m × ΔHfus = (25.0 g) × (6.02 kJ/mol) ÷ (18.02 g/mol) = 8.35 kJ
Then, we need to calculate the heat required to raise the temperature of the resulting water from 0 °C to the equilibrium temperature T:
q2 = m × C × ΔT = (275.0 g) × (4.18 J/g°C) × (T - 0 °C)
Since the system is in thermal equilibrium, we know that q1 + q2 = 0:
8.35 kJ + (275.0 g) × (4.18 J/g°C) × (T - 0 °C) = 0
Simplifying and solving for T:
T = 64.9 °C
Rounding to the nearest degree gives us an equilibrium temperature of 65 °C (choice C).

What Would Be the Appropriate Equilibrium Temperature of a System