Answer:
[tex]\huge\boxed{\sf Q = 5880\ Joules}[/tex]
Explanation:
Given:
m = mass = 40 g = 0.04 kg
c = specific heat constant of water = 4200 J/Kg°C
ΔT = change in temperature = T2-T1 = 60 °C - 25 °C = 35°C
Required:
Q = change in heat energy = ?
Formula:
Q = mcΔT
Solution:
Q = (0.04)(4200)(35)
Q = 5880 Joules
[tex]\rule[225]{225}{2}[/tex]
