posted by Anonymous .
86.3 g of water in a calorimeter have an initial temperature of 24 degrees C. The temperature increased to 33.1 degrees C when 54.3 g of a sample of metal initially at 99.7 degrees C is added to the calorimeter and allowed to equilibrate with the water.
Calculate the specific heat of the metal in joules/g degrees C.
qH2O + qmetal = 0
qH2O = mass H2O x specific heat water x (Tfinal-Tinitial).
qmetal = mass metal x specific heat metal x(Tfinal-Tinitial).
Just plug in the numbers and solve for specific heat metal.
No, it isn't 23.
is it 186.1?