Question

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

Answer 1

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