Final answer:
The final temperature of the coffee-water mixture is approximately 90°C after an ice cube of mass 8.0 g at 0°C is added to 130 g of coffee at 95°C. This is calculated by taking into account the heat of fusion and the specific heat capacity of water.
Step-by-step explanation:
To determine the final temperature of the coffee-water mixture after adding an ice cube of mass 8.0 g at 0°C to a cup of coffee at 95°C, we must consider the energy needed to melt the ice and the energy exchanged between the melted ice and the coffee. The specific heat capacity of water is 4.184 J/(g°C), and the heat of fusion of ice is 6.0 kJ/mol. Since the mass of ice is 8.0 g (or 0.008 kg), and knowing that the molar mass of water is approximately 18.015 g/mol, we can calculate the moles of ice as:
Moles of ice = Mass of ice / Molar mass of water
= 8.0 g / 18.015 g/mol
= 0.444 moles (approximately)
The energy required to melt this amount of ice (Qfus) can be calculated as
Qfus = Moles of ice × Heat of fusion
= 0.444 moles × 6.0 kJ/mol
= 2.664 kJ
This energy will come from the coffee, cooling it down. We then use the formula Q = mcΔT to calculate the change in temperature (ΔT) of the coffee, where m is the mass of the coffee, c is the specific heat capacity, and ΔT is the temperature change:
Q = mcΔT
2.664 kJ = 130 g × 4.184 J/g°C × ΔT
Since 2.664 kJ is equivalent to 2664 J, we get:
2664 J = (130 g × 4.184 J/g°C) × ΔT
ΔT = 2664 J / (130 g × 4.184 J/g°C)
ΔT = 4.90°C
Therefore, the final temperature of the coffee will be:
Final temperature = Initial temperature of coffee - ΔT
= 95 u00b0C - 4.90°C
= 90.10°C (approximately)
Since 90.10°C is closest to 90°C from the given options, the correct answer is 90°C.