Question

A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling

Answer 1

Answer:

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

$Q=mCp\Delta T=mCp(T_2-T_1)$

In such a way, solving for the final temperature $T_2$ we obtain:

$T_2=T_1+\frac{Q}{mCp}$

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

$T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C$

Thereby, answer is B. 58°C .

Regards.