Answer:
-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
