Answer:
Heat transferred, Q = 1542.42 J
Explanation:
Given that,
Mass of water, m = 30 grams
Initial temperature, [tex]T_i=25^{\circ} C[/tex]
Final temperature, [tex]T_f=12.7^{\circ} C[/tex]
We need to find the energy transferred. The energy transferred is given by :
[tex]Q=mc\Delta T[/tex]
c is specific heat of water, c = 4.18 J/g °C
So,
[tex]Q=30\ g\times 4.18\ J/g-^{\circ} C\times (12.7-25)\ ^{\circ} C\\\\Q=-1542.42\ J[/tex]
So, 1542.42 J of energy is transferred.
