Question #816d2

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Doc048
Jun 18, 2017

#Q = mcDelta T#

#"mass" (m) = 250.0 "grams"#
#"specific heat" (c) = 0.38 J/g ^oC#
#"Temperature change" (DeltaT) = (100 - 45)^oC = 55^oC#

#Q = 250.0gxx0.38 J/g^oCxx55^oC = 5,225 _"Joules"#

#"In"# #"calories"# = #(5,225 "Joules")/(4.184J/"calorie") = 1,249_" calories"#

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