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.

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