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What will the final temperature be if 688 calories of heat is released by 25 mL of water with an initial temperature of 80 degrees celsius?
Wiki User
∙ 13y ago
Do some converting first.
688 calories (4.184 Joules/1 calorie)
= 2878.592 Joules
25 ml of water = 25 grams
q(Jolules) = mass * specific heat * (Temp. final - Temp. initial)
2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C )
2878.592 Joiles = 104.5( Temp. Final) - 8360
11238.592 =104.5(Temp. Final)
107.55 Celsius Final Temperature ( call it 108 C )
Wiki User
∙ 13y ago
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