*Review Problem 6.058 A sample of copper was heated to 111.98 then thrust into 2
ID: 960084 • Letter: #
Question
*Review Problem 6.058 A sample of copper was heated to 111.98 then thrust into 200.0 g of water at 25.0o oc. C and The temperature of the mixture became 27.58 oc. Specific Heats Specific Heat, Jag c (25 C) Substance 0.7 Carbon (graphite) 0.387 Copper 2.45 Ethyl alcohol 0.129 Gold 0.803 Granite 0.4498 Iron 0.128 Lead 2.0 Olive oil 0.235 Silver 4.184 Water (liquid) How much heat in joules was absorbed by the water? the tolerance is +/-3% The copper sample lost how many joules? the tolerance is +/-3% sample? What is the heat capacity of the copper 1 J OC SCu the tolerance is +/-3% sample? What was the mass in grams of the copper mcu the tolerance is +/-3% Question Attempts: 0 Copyright C 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved. en.wileyplus.com/edugen/shared/assignment/test/qprint.unExplanation / Answer
Heat absorbed by water= mass* specific heat* temperature difference= 200*4.18*(27.68-25)=2240.48 joules
Heat abosrbed by water= heat lost by copper = 2240.48 joules
Heat abosrobed by water= heat capacity of copper* temperature difference= heat capacity*(111.98-27.68)= 2240.58
Heat capacity of copper = 2240.58/84.3=26.57 J/deg.c
Heat absorbed= mass of copper* specific heat of copper* temperature difference=
Mass of copper= 2240.58/(0.387*(111.98-27.68)=68.67 gms
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