You are making iced tea. The tea itself is initially at a temperature of +50.0 °
ID: 1464183 • Letter: Y
Question
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 g 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 105 J/kg. [4 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) (i) 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 representing the heat, in kilojoules, added to the ice and a negative bar 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: ____ ____
Explanation / Answer
a)delta_Q(water)=m*c*delta_T(water)=m*4000*(-20)=-m*80000
delta_Q(ice)=m*c*delta_T(ice)=m*2000*50
Because thermal energy is conserved in this mixing process
delta_Q=delta_Q(water)+delta_Q(ice)=0
==>-80000m+5000=0
==>m=0.0625kg=62.5g
b)1.temperature at which ice starts to melt=1 degrees
heat=.05*1*2000=100 J
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