Answer: - 452.088joule
Step-by-step explanation:
Given the following :
Mass of water = 12g
Change in temperature(Dt) = (11 - 20)°C = - 9°C
Specific heats capacity of water(c) = 4.186j/g°C
Q = mcDt
Where Q = quantity of heat
Q = 12g × 4.186j/g°C × - 9°C
Q = - 452.088joule