Exercise 17.44 In a container of negligible mass, 0.340 kg of ice at an initial

Question

Exercise 17.44

In a container of negligible mass, 0.340 kg of ice at an initial temperature of -27.0 C is mixed with a mass m of water that has an initial temperature of 80.0 C. No heat is lost to the surroundings.

Part A

If the final temperature of the system is 28.0 C, what is the mass m of the water that was initially at 80.0 C?

Express your answer to three significant figures and include the appropriate units.

Exercise 17.44

In a container of negligible mass, 0.340 kg of ice at an initial temperature of -27.0 C is mixed with a mass m of water that has an initial temperature of 80.0 C. No heat is lost to the surroundings.

Part A

If the final temperature of the system is 28.0 C, what is the mass m of the water that was initially at 80.0 C?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

heat of fusion ice = 334*10^3 J/kg
c of ice= 2100 J/kg K
c of water = 4190 J/kg K

At the final temperature .. heat gained by cooler mass = heat lost by warmer mass.

Heat gained by ice ..
• Ice at -28 to 0 .. (m x ci x ) = (0.34kg x 2100 x 28) = 1.999^4 J
• Melting ice .. (m x Lf) = (0.34 x 334^3) = 12.793^4 J
• Melted water 0 to +29 .. (m x cw x ) = (0.34 x 4190 x 28) = 3.988^4 J

•• total heat gain = (1.99 + 12.79 + 3.98)^4 J = 18.76^4 J

Heat loss by hot water = (Mw x cw x )
= Mw x 4190 x (80-28) = 2.178^5 .Mw

Equating the heat lost and gained ..
2.178^5.Mw = 18.76^4 J .. .. Mw = 18.76^4 / 2.178^5 .. .. .. Mw = 0.2527 kg

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

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