A 3.985 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.17 ∘C to 30.90 ∘C.

Question

The heat capacity (calorimeter constant) of the calorimeter is 43.07 kJ/ ∘C, what is the heat of combustion per gram of the material?

Answer 1

q = Ccal∆T = (43.07 kJ/º)(30.90 - 24.17) = (43.07 kJ/º)(6.73º) = 289.86 kJ

289.86 kJ/3.985 g = 72.74 kJ/gram

Answer 2

Hello, Adoria,

Set up a table for what you know, and need to find. Use the formula q = cmDT, where DT is (T1 - T2), the initial and final temperatures.

The energy is negative since heat is released from the reaction into the system (calorimeter). q is the total heat from the reaction, so divide that by the mass of the substance to provide units of kJ/gC.

Bob