Did I do this right?
q water = 24 ml x 4.18 J / (g ˚C) x (27.5 ˚C – 24 ˚C)
q = 351 J water = -351 J unknown metal
-351 J = 15.262 g x c x (27.5 ˚C – 100.3 ˚C)
c = -351 / (15.262 x -72.8)
c = 0.316 J (g ˚C)
Mass of metal: 15.262 g
Volume of water in the calorimeter: 24.0 mL
Initial temperature of water in calorimeter: 25.2 °C
Temperature of hot water and metal in hot water bath: 100.3 °C
Final temperature reached in the calorimeter: 27.5 °C

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