Answer:
Q = -22.9 kJ
Explanation:
Given that,
Mass of water, m = 150.3 g
Water gets cool from 25.60°C to -10.70°C.
The specific heat of water, c = 4.2 J/g°C
The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=150.3\times 4.2 \times (-10.7-25.6)\\\\Q=-22914.738\\\\or\\\\Q=22.9\ kJ[/tex]
So, 22.9 kJ of heat is needed to be removed to cool.
