Answer:
-3.617 °C
Step-by-step explanation:
Step 1: Given data
Mass of water (m): 210.0 g
Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)
Specific heat of water (c): 4.184 J/g.°C
Temperature change (ΔT): ?
Step 2: Calculate the temperature change
We will use the following expression.
Q = c × m × ΔT
-3178 J = 4.184 J/g.°C × 210.0 g × ΔT
ΔT = -3.617 °C