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