The question is: Calculate the heat capacity calorimeter at J / K. The information given to me: A coffee cup calorimeter containing 140th 0g water at 25. 1degrees C. A 123rd 0-g block of copper heated to 100. 4 degrees C. The specific heat of Cu (s) is O. 385 J / GK. The final temperature of the contents of the cup is 30. 3 degrees C. I found: q (Cu) = 3320Jq (H2O) = 3040 J
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i got 3057 J for the water (rounded)
and 3319.6J for the copper
where did the extra 280J go though? it couldn’t have just disappeared… (law of conservation of energy), it was absorbed by the room temperature calorimeter! Heat capacity is the amount of energy that can be stored in an object, in this case, 280J.
ps: Thanks for organizing your question well. too many people list “PLZ HLP” in their question and forget to list the other information given.