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