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 the 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

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter16: Temperature And The Kinetic Theory Of Gases
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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 the 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
6.
9
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 the 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 6. 9
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Step 1

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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