Question #816d2

1 Answer
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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