A customer orders 200ml of Sumatran coffee at precisely 60.0°C. You then need to drop the temperature of the coffee, initially at 90.0°C, to the ordered temperature. In order to simplify the calculations, you will start by assuming that coffee has the specific heat and density as if water. In the following parts, you will remove these simplifications. Solve now this problem assuming the density is 1.000 g/ml for coffee and its specific heat capacity is 4.184 J/(g ºC). If you had used ice cubes to cool the coffee, your calculation of q would have been a two- step process: (1) the ice at 0 oC first has to melt (energy added to go from solid to liquid; and then (2) to warm the liquid from 0 oC to the final temperature where  q = mice cice     +     mwater cwater ∆T          where  cice =   2.11 J/ g . oC     cwater =  4.184 J/ g . oC  (Note: there is no ∆T for the first step since melting of the ice occurs at 0 oC,  no temperature change) What mass of ice would you need to use to achieve the same result (qice = qmilk)?  (Note: mice = mwater)

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A customer orders 200ml of Sumatran coffee at precisely 60.0°C. You then need to drop the temperature of the coffee, initially at 90.0°C, to the ordered temperature.

In order to simplify the calculations, you will start by assuming that coffee has the specific heat and density as if water. In the following parts, you will remove these simplifications. Solve now this problem assuming the density is 1.000 g/ml for coffee and its specific heat capacity is 4.184 J/(g ºC).

If you had used ice cubes to cool the coffee, your calculation of q would have been a two- step process: (1) the ice at 0 oC first has to melt (energy added to go from solid to liquid; and then (2) to warm the liquid from 0 oC to the final temperature where

 q = mice cice     +     mwater cwaterT          where  cice =   2.11 J/ g . oC

    cwater =  4.184 J/ g . oC

 (Note: there is no T for the first step since melting of the ice occurs at 0 oC,  no temperature change)

What mass of ice would you need to use to achieve the same result (qice = qmilk)?  (Note: mice = mwater)

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