Answer:
Q = -811440 J
Explanation:
Given data:
Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)
Initial temperature = 191 °C
Final temperature = 23°C
Specific heat capacity of oil = 1.75 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 23°C - 191 °C
ΔT = -168°C
Q = 2760 g ×1.75 J/g.°C ×-168°C
Q = -811440 J
Negative sign show heat is released.
