If 2500. j of energy are added to 120. g of benzene at 30. degrees c, what will be its final temperature?

When heat (q) is absorbed by "m" grams of a substance then the change in temperature is given as,

                                         q  =  m Cp (T₂ - T₁)  --- (1)

where;
           Cp  =  Specific Heat

Specific heat of Benzene is 1.72 J/g.°C

Now,
Solving equation 1,

                             (T₂ - T₁)  =  q / m Cp
Putting values,
                             (T₂ - 30 °C)  =  2500 J ÷ (120 g ×1.72 J/g.°C)

                             (T₂ - 30 °C)  =  2500 J ÷ (206 J/°C)

                             (T₂ - 30 °C)  =  12.13 °C

                             T₂  =  12.13 °C + 30 °C

                             T₂  =  42.13 °C

257 grams h2o i think