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 sur

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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 sur

roundings, 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

Chemistry

2 answers:

KengaRu [80] · Answer 1 · 2022-02-27T06:00:16+03:00

7.0 Â°C
We have 40.0 g of ice at -11.0Â°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 0.0Â°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 Â°C + -16.64807437 Â°C = 8.351925631 Â°C
We now effectively have 2 masses of water. A 40 gram mass at 0 Â°C and a 205 gram mass at 8.351925631 Â°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 Â°C