A calorimeter contained 75g of water at 16.95C. A 93g sample of iron at 65.58C was placed in it, giving a final temperature of 19.68C for the system. Calculate the heat capacity for the calorimeter. Specific heats are 4.184 for water and 0.444 for Fe.
The answer is suppose to be 381J/C but I dnt know how it was derived, can anyone explain the answer?
2007-08-07 17:43:50 · 1 answers · asked by Mr S 2 in Science & Mathematics ➔ Chemistry
The usual equation is
Heat lost from iron= Heat gained by water and calorimeter. For each item, heat= weight x SpecHeat x Change in temperature.
Apparently you should compute the loss of heat from the iron, which will be greater than the gain by the water. The remainder is attributable to the calorimeter, and the weight x specific heat are combined to come up with the 381 joules/degC.
2007-08-07 18:29:13 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋