A 256-g sample of an unknown substance was heated from 15.0 °c to 65.0 °c. In the process, the substance absorbed 7369 j of energy. What is the specific heat of the substance?

Answer 1

For this question you would start with the equation, Q = mcΔT, where Q is heat in joules (J), M is mass of the substance in grams (g), c is the specific heat in grams per joule degree Celsius (g/J °C).

To start out with, write out the equation that you are going to use and all of the information that you have so far, so that you don't have to keep rereading the question to find it.

Q = mcΔT where:

Q = 7369J

m = 256g

C = ?

ΔT = 65.0^°C - 15.0^°C = 50^°C

Because you are solving for specific heat (c), you will then rearrange the equation above so that you are solving for specific heat (c) instead.

c = Q ÷ (mΔT)

From here you just plug in the numbers that your recording above in there proper places and solve.

c = 7369J ÷ (256g × 50^°C)

c = 7369J ÷ 12800g^°C

c = 0.576g/J^°C