CH301 - iceteaprob

CH301 - iceteaprob - Iced-Tea Problem Its hot and I want a...

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Iced-Tea Problem 220 g of 35˚C tea (water) is poured over 250 g of -10˚C ice. What happens. .. The warm water (tea) will cool down to 0˚C. As it cools the ice will heat up to 0˚C and then start to melt. There is plenty of ice however and only a portion of it will melt. Find the amount of ice that melts into water and you’ll know the final composition and even the dilution factor on the tea. Cool tea water down to 0˚C 4.184 J/g˚C · 220 g · 35˚C = 32216.8 J removed to cool tea to 0˚C Warm ice up to 0˚C 2.09 J/g˚C · 250 g · 10˚C = 5225 J to heat ice up to 0˚C This heat comes from the cooling water, so substract this amount off the total above 32216.8 - 5225 = 26991.8 J of heat that will MELT the ice mass of ice melted = 26991.8 / 334 = 80.8 g of ice melted 220 g + 80.8 = 300.8 g of water 250 g - 80.8 = 169.2 g of ice final mixture for my glass of iced tea all at 0˚C 220.0 g original tea 300.8 g final tea The dilution factor is simply the original volume over the final volume (or masses here ).
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