Two 20.0-g ice cubes at –18.0 °C are placed into 215 g of water at 25.0 °C. Assu

Question

Two 20.0-g ice cubes at –18.0 °C are placed into 215 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.

Two 20.0-g ice cubes at-18.0 are placed into 215 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 H20(s) 37.7 J(mol K) heat capacity of H20() 75.3 J/(mol K) enthalpy of fusion of H20 6.01 kJ/mol Number

Explanation / Answer

Answer

Tf = 7.15

Explanation

Heat released by water(q(rel) =- Heat absorbed by ice(q(obs))

Heat abserbed by ice = Heat required to raise the ice cubes from -18 to 100(q1) + Heat required to fuse ice cubes(q2) + Heat required to bring the temperature of melted iceto final value(q3)

so,

q(rel) = - (q1 + q2 + q3)

q(rel) = m× T × C

= 215g × (X - 298.15K) × 4.814J/g K

= X899.6J/K- 268216J

q1 = m × T × C

= 40g × 18K × 2.094J/g K

= 1507.7J

q2 = (6010J/18g) ×40g = 13356J

q3 = 40g × (x - 273.15K ) × 4.184J/g K

= X167.4J/K- 45725J

X899.6J/K - 268216J = -(1507.7J + 13356J + X167.4J/K - 45725J)

X1067J/K = 299077J

X = 280.3K = 7.15

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.