Answer:
The final temperature of the water is 6.53 °C
Step-by-step explanation:
Step 1: Data given
Mass of ice cubes = 2*20.0 grams = 40.0 grams
Temperature of the ice = -17.0 °C
Mass of water = 205.0 grams
Temperature of water = 25.0 °C
Heat Capacity of H2O(s) = 37.7 J/(mol.K)
Heat Capacity of H2O(l) = 75.3 J/(mol.K)
Enthalpy of fusion of H2O 6.01 kJ/mol
Step 2: Calculate moles of H2O
Moles H2O = mass H2O / molar mass
40.0 g H2O(s) / 18.02 g/mol = 2.22 mol H2O
Step 3: Calculate energy needed to warm the ice to melting temperature
(37.7 J/(mol*K)) * (2.22 mol) * (0 - (-17.0))K = 1422.8 J
Step 4: Calculate energy needed to melt the ice
(6.01 KJ/mol) * (2.22 mol) = 13.34 kJ = 13340 J to melt the ice
Step 5: Calculate total energy to warm and melt all the ice
1422.8 + 13340 J = 14762.8 Joule
Step 6: Calculate moles of water
Moles H2O = 205 grams / 18.02 g/mol
Moles H2O = 11.38 mol
Step 7: Calculate the change in temperature
ΔT = (14762.8 J / 75.3 J/mol*k) /11.38 mol
(15101.8 J) / (75.3 J/(mol*K) / (205 g / (18.01532 g H2O/mol)) = 17.2
Step 8: Calculate new temperature of water
25.0 - 17.2 = 7.8 °C
We have 2 bodies of water: 40 g at 0°C and 205 g at 7.8°C
⇒ ((40 g * 0°C) + (205 g * 7.8°C)) / (40 g + 205 g) = 6.53 °C
The final temperature of the water is 6.53 °C