Class Management i Help Problem Set 2 - Vol 2 Ch 1 & 2.1 - Temperature, He Begin

Question

Class Management i Help Problem Set 2 - Vol 2 Ch 1 & 2.1 - Temperature, He Begin Date: 2/3/2017 12:00:00 AM -- Due Date: 9/19/2017 11:59:00 PM End Date: 12/15/2017 11:59:00 PM (10%) Problem 10: You mix m1 = 0.65 kg of ice at T1 =-18°C with mw = 4.8 kg of water at Tw = 70°C in an insulated container. The specific heats of ice and water are cl = 2.10x103 J/(kg·°C) and cw = 4.19×103 J/(kg·°C), respectively, and the latent heat of fusion for water is Lf = 3.34 × 105 J/kg LS 50% Part (a) Enter an expression for the final, equilibrium temperature of the mixture, in terms of the defined quantities Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 5 Cw % per attempt) detailed view mw 0 BACKSPACE Submit Hint Hints: 1 % deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. D 50% Part (b) Calculate the final, equilibrium temperature of the mixture, in degrees Celsius

Explanation / Answer

energy required by ice as its temp increases to 0 deg C,

Q1 = m C deltaT = (0.65) (2.10 x 10^3) (0 - (-18))

Q1 = 24570 J

Energy Required to melt the ice,

Q2 = m Lf = 0.65 x 3.34 x 10^5 = 217100 J

supose final equilibrium temp is T,

energy Required, Q3 = (0.65)(4.19 x 10^3) (T - 0 )

Q3 = 2723.5 T

energy released by water,

Q4 = 4.8 x 4.19 x 10^3 ( 76 - T )

Q4 = 1528512 - 20112 T

Applying energy conservation,

Q1 + Q2 + Q3 = Q4

- m1 c1 T1 + m1 c1 + m1 cW T = mW cW TW - mW cW T

T = (mW cW TW - m1 c1 + m1 c1 T1) / (mW cW + m1 cW )

(B) 24570 + 217100 + 2723.5 T = 1528512 - 20112 T

T = 27 deg C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---