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5.2 Heat and Heat Capacity Understand how to use spedfc heat in heat loss/ gain caloulations Question The specific heat of wa

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Answer #1

Given:

Q = 59000 J

m = 600 g

C = 4.184 J/g.oC

Ti = 75.5 oC

use:

Q = m*C*(Tf-Ti)

59000.0 = 600.0*4.184*(Tf-75.5)

Tf -75.5 = 23.5 oC

Tf = 99.0 oC

Answer: 99.0 oC

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