Answer
ΔT = 3.78x10^-3 °C
Step-by-step explanation
Given:
Energy (Q) = 12.0275 J
Mass of water = 760.6 g
We know the specific heat capacity of water = 4.182 J/g °C
Required: The change in temperature
Solution
Q = m x Cp x ΔT
ΔT = Q/m x Cp
ΔT = 12.0275 J/(760.0 g x 4.182 J/g °C)
ΔT = 3.78x10^-3 °C (or 0.00378 °C)