Question

PROBLEM 5-8 points You are making iced tea. The tea itself is initially at a temperature of 50.0 °C. You then toss in an ice cube with a mass of 50 and an initial temperature of -20.0 °C. The ice cube melts, and the final temperature of the mixture is +30.0 °C. For this problem, use the following values: c for liquid water is 4000 J/(kg "C), c for solid water is 2000 J/(kg°C), and the latent heat of fusion of water is 3.0 x 10 J/kg. [2 points] (a) Assuming no heat is exchanged with the surroundings, calculate the mass of tea that you started with Now, we'll break the problem down into three parts. [3 points] (b) Step 1 - bringing the ice to the temperature at which it melts. How much heat does this require and by how much does this cool the tea? (ii) Step 2 - Melting the ice. How much heat does this require and by how much does this cool the tea? (iii) Step 3 - Bringing the melted ice to the final temperature. How much heat does this require and by how much does this cool the tea? [3 points) (c) Plot three pairs of bar charts, corresponding to the three steps above. For each step, show a positive bar (up) representing the heat, in kilojoules, added to the ice and a negative bar (down) representing the heat, in kilojoules, removed from the tea. The bars should be vertical. Use the same scale for every case, so that you display visually how the heats compare to one another. STEP 1: STEP 2: STEP 3:

Answer 1

solution Gioen data liquid water - 9000 J/(kg°C) Solid water = 2000 J/K9°C Catent heat = 3 x10$J/kg (a) now, let The mass of the team total heat lost water & heat gained by Then, - ice MCAT = mice Cice OT + micel tmice Cot m/4000) (50-30) -(50x10% kg) Cocoog (20) + C50X10") (3x105)+ (50x103) (4000) (30C ml socco) . 2000 + 15000 76000 m = 23000 80000 = 0.2875 kg - 887.5 grams m (6 af bring the ice to melting point. Then, the heat required to being the final temperature Q - (50 x10°3) (2000) (20°C) Q - 2000 J MCAT - mice Cice at Ics Scanned with CamScanner

(0.3875) (4000) (50 - 7) = (50x103) (2000) (Qo'c) 0.28 75 X 4000 x(50-7) - 2000 6 -7 - 2cao toc xon87 50 -T 1.74 I 50 -1.74 IT - 48.9°C/ © If melting the ice. Then, the heat required to bring the pinal temperature Q = (50x1033) (2000) (20°) +(50x1013) (3x103) - 17000 J Now, calculating the final tempesature of the tea. (0.2875) (4000) (50 - 7) = 2000 + 15000 50-7 = 17000 02875X1000 50 -T 14.78 T = 50-14078 T = 35°c) calculating heat se quired to bring the d) Now, Final temperature 18 Q = (50x103) (2000) (20) + (50x103) (3x105) + (50x103) (4000) (30°C) Canne CamScanne

♡ - 23000T is, final temperature (0.2875) (4000) (50-7) - 23000 50 -7 = 23000 4000/ 0-2875 T = 50- 23000 4000X 0.2875 IT = 30°c) CamScanne