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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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
Transcribed Image Text:Before brewers had access to propane or natural gas, they would heat large, insulated vats of water by adding hot stones to the water. The stones are heated to 800°C before they are dropped into a particular well-insulated large vat holding 1,200 kg of water at 20°C. If we want to heat the water to 70°C, what mass of stones must be added to the vat? The specific heat of the stone is about 790 J/kg°C and the specific heat of water is 4,186 J/kg°C. Assume the mass of the water does not change and that the vat is a perfect insulator so no internal energy is lost to the surroundings.
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Heat gained by water = mass of water * specific heat of water * change in temperature = 1200 * 4186 * (70-20) = 251160000 JoulesNow, this amount of energy must be lost by stone Thus , Hstone = - 251160000 = mass of stone * specific heat of stone * change in temperature of stone -251160000 = - m * 790 * (70 - 800)m = 435.5 kg
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