Answer to Question #90987 in General Chemistry for ruzzle harvey

Question

a calorimeter contained 75.0g of water at 16.95 degress celcius. a 93.3 g sample of iron at 65.58 degrees celcius was placed in it, giving a final temperature of 19.68 degres celius for the system. calculate the heat capacity of the calorimeter specific heats are 4.184J/g/degrees celcius for h20 and 0.444 J?g?degrees celicius for Fe

Answer 1

$\Delta t=$ 65.58 - 19.68 = 45.9^oC

$q=mc\Delta t =93.3g*45.9C*0.444 J/g*C=1901.42 J$

$\Delta t(H2O)= 19.68 - 16.95 = 2.73$^oC

$q=75g*4.184J/g*C*2.73C=856.674 J$

1901.42 J - 856.674 J = 1044.746 J

$c=q/ \Delta t =1044.746 J/2.73$^oC=382.69 J/^oC