Answer:
[tex]T_f=7.38^{\circ}[/tex]
Explanation:
Given that,
Mass, m = 5 kg
Initial temperature of water, [tex]T_i=5^{\circ} C[/tex]
Heat transferred, Q = 50 kJ = 50,000 J
The specific heat capacity of water is 4200 J/kg°C
We need to find the final temperature of the water. The heat transferred in the process is given by :
[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\\dfrac{Q}{mc}+T_i=T_f\\\\T_f=\dfrac{50000}{5\times 4200}+5\\\\T_f=7.38^{\circ}[/tex]
So, the final temperature of water is [tex]7.38^{\circ}[/tex].
