1. Iced tea is made by adding ice to 1.8 kg of hot tea, initially at 80°C. How m
ID: 1702032 • Letter: 1
Question
1. Iced tea is made by adding ice to 1.8 kg of hot tea, initially at 80°C. How many kg of ice, initially at 0°C, are required to bring the mixture to 10°C? (Lf = 3.33 x 105 J/kg, cw = 4 186 J/kg·°C)a. 1.8 kg
b. 1.6 kg
c. 1.4 kg
d. 1.2 kg
2. A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C. What is the final equilibrium temperature? (Specific heat for aluminum is 900 J/kg·°C, the specific heat of water is 4 186 J/kg·°C, and Lf = 3.33 × 105 J/kg.)
a. 17.9°C
b. 9.5°C
c. 12.1°C
d. 20.6°C
Explanation / Answer
The mass of hot tea mtea = 1.8kg
the initial temperature Ti = 80
the final temperature of the system is Tf = 10
From calorimetry
mtea c (80 -10) = mice L + mice c (10)
Therefore the mass of the ice
mice = mtea c (70) / [ L + c(10)]
= (1.8)(4186)(70) / [(333*10^3 + (4186)(10)]
= 527436 / 374860
= 1.4kg
(2) The mass of ice mice = 0.05kg
The mass of water m = 0.2kg
The mass of the aluminium mAl = 0.08kg
The initial temperature T = 30
Specific heat for aluminum is 900 J/kg·°C,
the specific heat of water is 4 186 J/kg·°C,
and Lf = 3.33 × 105 J/kg
From calorimetry
(0.05)(3.33*10^5) + (0.05)(4186)(T) = (0.2)(4186)(30 - T) + (0.08)(900)(30-T)
16650 + 209.3T = 25116 - 837.2T +2160 - 72T
1118.5T = 10626
then T = 9.5deg
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