Question

Initially you have m_W = 3.4 kg of water at T_W = 54°C in an insulated container. You add ice at T_I = -21°C to the container and the mix reaches a final, equilibrium temperature of T_f = 25°C. The specific heats of ice and water are c_I = 2.10×10³J/(kg⋅°C) and c_W = 4.19×10³ J/(kg⋅°C), respectively, and the latent heat of fusion for water is L_f = 3.34×10⁵ J/kg.

(11%) Problem 7: Initially you have mw = 3.4 kg of water at Tw = 54°C in an insulated container. You add ice at Ti = -21°C to the container and the mix reaches a final, equilibrium temperature of Tp = 25°C. The specific heats of ice and water are 1 = 2.10x103 J/(kg:°C) and cw = 4.19x109 J/(kg:°C), respectively, and the latent heat of fusion for water is Lp = 3.34x10 J/kg. A 50% Part (a) Enter an expression for the mass of ice you added, in terms of the defined quantities. m = 1 Grade Summary Deductions 0% Potential 100% B Y HOME 1 Cw h 0 d Lp n TW 0 7 8 4 5 / * 1 2 + - 0 VO BACKSPACE 9 6 3 . DEL Submissions Attempts remaining: 50 (0% per attempt) detailed view m mw END CLEAR T Submit Hint Feedback I give up! Feedback: 0% deduction per feedback. Hints: 0% deduction per hint. Hints remaining: 3 A 50% Part (b) Calculate the mass of ice you added, in kilograms.

Answer 1

Given mw = 3-4 kg Tw = 54 C To = -21°C If = 250 C: = 2.1 x 1 /-k Cu = 4.19 110°J-va-c L = 2.34 x 6 / a) Recall Principle of calorimetry. a lost = again 29 mw Cw ☆ MwCwX2q = MICEX21 +MILE + m2 CwX25 = MI = 2161 ++ 2560 3.4X 4 ola x 10x154–25) = Mpx 2.1x10x21 + MAX 3.3 4x10 + m2X4•19 xlox(2) → 413134 = m;44100 + 334000 + 104750] m = 0.8556 kg ~ 0.86k