Question

Two 20.0-g ice cubes at –11.0 °C are placed into 205 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.
Heat capacity of H2O(l)= 75.3J/(mol*K)
Heat capacity of H2O(s)=37.7J/(mol*K)
Enthalpy of Fusion of H2O=6.01kJ/mol

Answer 1

7.0 °C  
We have 40.0 g of ice at -11.0°C. Since our heat capacity constants are per mole, let's convert all our measurements to moles. Start by calculating the molar mass of H2O 
Atomic weight oxygen = 15.999 
Atomic weight hydrogen = 1.00794  
Molar mass hydrogen = 15.999 + 2 * 1.00794 = 18.01488 g/mol 
Moles ice = 40.0g / 18.01488 g/mol = 2.220386703 mol 
Moles water = 205 g / 18.01488 g/mol = 11.37948185 mol  
Let's first calculate how much energy needs to be absorbed from the environment to get that ice to a temperature of 0.0°C. 
2.220386703 mol * 11 K * 37.7 J/(mol*K) = 920.7943655 J  
Now let's melt that ice. 
2.220386703 mol * 6010 J/mol = 13344.52408 J  
So the total energy to melt the ice is: 
13344.52408 J + 920.7943655 J = 14265.31845 J  
Now calculate the amount of mole degrees that had to be extracted to get that much energy. 
-14265.31845 J / 75.3 J/(mol*K) = -189.4464601 mol*K  
Divide by the amount of water being cooled. 
-189.4464601 mol*K / 11.37948185 mol = -16.64807437 K  
So the temperature dropped by a bit under 17 degrees K. So 
25.0 °C + -16.64807437 °C = 8.351925631 °C  
We now effectively have 2 masses of water. A 40 gram mass at 0 °C and a 205 gram mass at 8.351925631 °C. Obviously that isn't stable, so we need to bring both masses to the same average temperature. So let's do a weighted average: 
(40 * 0 + 205*8.351925631)/(40+205) 
= (0 + 1712.144754)/245 
= 6.988345936 
 Rounding to 1 decimal place gives a final temperature of 7.0 °C

Answer 2

The final temperature, T, of the water after all the ice melts = 6.9 °C

Further explanation

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Two 20-g ice cubes = 40-g / 18.0-g / mol = 2.22 mole of ice

205-g / 18.0g / mol = 11.39 mole water

• heat to raise 40 g ice from -11 to zero.

Q1 = 2.22 x 37.7 J / (mol ° C) x 11

Q1 = 920.63 J

• heat to melt 40 g ice.

Q2 = mole x heat fusion

Q2 = 2.22 x 6.01kJ / mol

Q2 = 13.34 kJ = 13340 J

• heat to raise melted water to final T.

Q3 = 2.22 x 75.3J / (mol °C x (T-0)

Q3 = 167,166 T

• heat absorbed by 205 g H2O

Q4 = 11.39 x 75.3J / (mol ° C x (25-T)

Q4 = 857.67 (25-T)

Then:

Q1 + Q2 + Q3 = Q4

920.63 J + 13340 + 167,166 T = 857.67 (25-T)

14260.63 + 167.166T = 21441.75 - 857.67T

857.67T + 167.166T = 21441.75 - 14260.63

1027.84 T = 7181.12

T= 6.98 C

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Keywords: heat, temperature