A 100.0g sample of aluminum is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150g of water at 25.0 oCAfter the metal cools, the final temperature of the metal and the water is 34.1 oCCalculate the specific heat capacity of aluminum from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeterSpecific heat of water 4.184 J/g oCatomic mass Al 27.0 g/molCalculate your answer in J/g oC with 3 significant figures, but enter the answer without units.Then calculate the molar heat capacity of Al in J / mol oC, again, to 3 significant figuresEnter the answer without units.The heat given off from aluminum (from initial heating) will eventually equal the heat energy absorbed by the water at the final temperature, 34.1 oCqAl qWater q m.c.deltaT (100.0g Al)(X J/g oC)(100 - 34.1) (150.0g H2O)(4.184 J/goC)(34.1 - 25.0) Solve for X, specific heat capacity of aluminum 6590X 5711.16 X 0.867 J/g oC Al The molar heat capacity in J/mol oC Molar mass x specific heat Al 27.0g/mol x 0.867 J/g oC grams cancel, leaving J/mol oC 23.4 J/mol oCOther related question