Two 20.0-g ice cubes at -17.0 c are placed into 205 g of water at 25.0c. assuming no energy is transferred to or from the surroundings. calculate the final temperature of the water after all the ice melts. 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

The final temperature of the water is 6.53 °C

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

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

what is the question here. i think you messed up.