A burner on an electric range has a heat capacity of 345 J/c. what value of q, in kJ, is produced when the burner cools from a temperature of 467 C to room temperature of 23 C?

-153.18 kj

Explanation:

Given dа ta:

Heat capacity of burner = 345 j/°C

Initial temperature = 467°C

Final temperature = 23°C

Heat produced = ?

Solution:

Formula:

q = cΔT

q = heat absorbed or released

c = heat capacity

ΔT = change in temperature

ΔT = 23°C - 467°C

ΔT = - 444°C

q = cΔT

q = 345 j/°C * - 444°C

q = - 153180 J

joule to kj:

q = - 153180 J * 1 kj / 1000 j

q = - 153.18 kj