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, ΔHfusion = 6.02 kJ/mol.]

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to melt the ice:
q_ice = (25.0 g) x (6.02 kJ/mol) / (18.02 g/mol) = 8.344 kJ

Next, we need to calculate the amount of heat lost by the coffee as it cools down:
q_coffee = (250.0 mL) x (0.997 g/mL) x (4.18 J/g°C) x (80°C - T)

Since the heat lost by the coffee is equal to the heat gained by the ice, we can set the two equations equal to each other and solve for T:
8.344 kJ = (250.0 mL) x (0.997 g/mL) x (4.18 J/g°C) x (80°C - T)

Simplifying the equation, we get:
T = 65°C

What Would Be the Appropriate Equilibrium Temperature of a System