Question

home study scence chemistry general chemistry / general chemistry solutions manuals/ chemistry 10th edition / Chapter 5 problem 190 Chemistry (0th Edition) Soe this solution in the app Chapter 5 Problem 19E Show all steps: OFF Proble Calorimetry How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver Spoon (specific heat 0.24 J/g °C) at 25 C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water Step-by-step solution Step 1 2 The amount of heat produced or absorbed by a sample is related to its specific heat as shown in the following formula qcxmxAT

Answer 1

We know that [q = mc AT gain where q= heat loss on heat m= mass c= specific heet AT= change in temperature Now, coffee = heat Heat loss by cup of Cinitially at 95°c) gained by silven spoon. Cintially at 25°C) - 1m, c, AT, = M₂ Cz DT2 ) - 0 where lis 2 is for for coffee silver spoon. - Using the given velues from the question. mi = 180g m2 = 45g C = 4.18 J/g c C2 = 0.245/9°C DTi = (95-x)°C where x is find tempereture, since heat is lost by cup of coffee, it will be smaller than initial temperahue AT2 a (x-25°C, where x is find temperature.

Putting these values in ☺ we get 180 g * 4.18 J/gºcx (95-x) C = 459 x 0.247/gºC x(22-25)°C = 752.4 J/°C X (95->)°C = 10.85/C x (x - 250°C = 10.8x - 270 71478 - 752.4x 71748 = 763.2x x= 94.000 This is the finel temperature attained Thus the temperature reduced will be Onitial Hempeletere - Find temperature 295°C -94°C 21°C Thes, 1°C temperature is reduced
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