Answer:
[tex]T_f=18.641^{\circ} C[/tex]
Explanation:
Given that,
Heat raised, Q = 9400 J
Mass, m = 3.5 kg
Initial temperature of water, [tex]T_i=18^{\circ} C[/tex]
The specific heat of water, [tex]c=4186\ J/kg^{\circ} C[/tex]
We need to find the final temperature of water. The heat required to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\\dfrac{Q}{mc}=(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{9400}{3.5\times 4186}+18\\\\=18.641^{\circ} C[/tex]
So, the final temperature is [tex]18.641^{\circ} C[/tex].
