Answer:
Q = -5913.6 J
Explanation:
Given that,
Mass of the sample, m = 44 g = 0.044 kg
Initial temperature = 18°C
Final temperature = -14°C
We need to find the energy involved in this change. The energy involved in this process is given by :
[tex]Q=mc\Delta T[/tex]
c is specific heat of water
[tex]Q=0.044\times 4200\times (-14-18)\\\\Q=-5913.6\ J[/tex]
So, 5913.6 J of energy is involved in this change.
