Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]
In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:
[tex]T_2=T_1+\frac{Q}{mCp}[/tex]
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]
Thereby, answer is B. 58°C
.
Regards.
