Question

A mass of 34.05 g of H2O(s) at 273 K is dropped into 185 g of H2O(l) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi- librium has been reached. Assume that CP, m for H2O is con- stant at its values for 298 K throughout the temperature range of interest.

Answer 1

Answer:

The temperature of the system once equilibrium is reached, is 292 Kelvin

Explanation:

Step 1: Data given

Mass of H2O = 34.05 grams  

⇒ temperature = 273 K

Mass of H2O at 310 K = 185 grams

Pressure = 1 bar = 0.9869 atm

Step 2: Calculate the final temperature

n(ice)*ΔH(ice fusion) + n(ice)*CP(H2O)(Tfinal- Ti,ice) + n(H20)*CP(H2O)*(Tfinal-Ti,H2O) = 0

Tfinal = [n(ice)*CP(ice)*Ti(ice) + n(H2O)*CP(H2O)*Ti(H20) - n(ice)*ΔH(ice fusion)] / [n(ice)*CP(ice) +n(H2O)*CP(H2O)]

⇒ with n(ice) = moles of ice = 34.05 grams / 18.02 g/mol = 1.890 moles

⇒ with CP(ice) = 75.3 J/K*mol

⇒ with Ti(ice) = the initial temperature of ice = 273 K

⇒ with n(H2O) = the moles of water = 185.0 grams / 18.02 g/mol = 10.27 moles

⇒ with CP(H2O) = CP(ice) = 75.3 J/K*mol

⇒ with Ti(H2O) = the initial temperature of the water = 310 K

⇒ with ΔH(ice, fusion) = 6010 J/mol

Tfinal = [1.890 moles * 75.3 J/K*mol * 273 + 10.27 mol * 75.3 J/K*mol * 310 K - 1.890 moles * 6010 J/mol] / [1.890 moles *75.3J/k*mol + 10.27 mol * 75.3 J/K*mol]

38852.541 + 239732.61  - 11358.9 = 267226.251

Tfinal= 291.8 ≈ 292 Kelvin

The temperature of the system once equilibrium is reached, is 292 Kelvin