When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 98°C. What is the specific heat of olive oil?

The answer to your question is Cp = 1.66 J/g°C

Explanation:

Data

Heat = Q = 435 J

Mass = m = 3.4 g

Temperature 1 = T1 = 21°C

Temperature 2 = T2 = 98°C

Specific heat = Cp = ?

Process

1.- Write the formula of heat

    Q = mCp(T2 - T1)

-Solve for Cp

     Cp = Q / m(T2 - T1)

2.- Substitution

     Cp = 435 / 3.4(98 - 21)

3.- Simplification

      Cp = 435 / 3.4(77)

      Cp = 435 / 261.8

4.- Result

      Cp = 1.66 J/g°C