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 data:

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

the answer your looking for is c).

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.