(EGY) Calculating Specific Heat Capacity Lesson

Calculating Specific Heat Capacity

In order to learn how to calculate specific heat, it is first important to become familiar with the terms associated with specific heat. You'll need to know how to recognize the symbol for each term and understand what it means. Here are the terms that are commonly used in the equation for calculating the specific heat of a substance:

Specific Heat Equation Q (heat added) = cmÅT c- specific heat m - mass ÅT - change in temperature

  • The amount of heat energy is represented by "Q". The amount of heat is represented by "J", or Joules.
  • Specific heat is represented by "c" and is a specific constant for varying substances.
  • The mass of the sample is represented by "m".
  • Delta, or the "Δ" symbol, represents the change in a variable.
  • "T" is the temperature of the substance.
  • "ΔT" is determined by subtracting the final value (T2) from the initial value (T1).   This is written as ΔT = T2 - T1.
    • For example, if your first temperature (T1) is 20ºC, and your second temperature (T2) is 150ºC, then ΔT = 150ºC - 20ºC, or 130ºC.
    • Sometimes we will get a negative value for ΔT.   For example, if (T1) is 75ºC, and (T2) is 33ºC, then ΔT = 33ºC - 75ºC = -42ºC.   This negative ΔT just means that the temperature decreased.   For this class, you can disregard the negative sign to do the rest of your calculations.   But, do know that the negative or positive sign tells you what happened to the temperature and ultimately what happens to the energy!

You are not required to memorize the formula as it is listed on the Physical Science Formula Sheet that you can use on all quizzes and tests for this course.  However, you do need to be able to manipulate the formula should you be required to find the specific heat instead of the energy required.

mage showing the equations "Q = cmΔT" and "c = Q / mΔT" with an explanation of isolating c.

Steps for Calculating Specific Heat:

Let's use the following problem to walk through the necessary steps:

Find the specific heat of 0.35 kg of an unknown substance when 34,700 J of heat is applied, and the temperature rises from 22 ºC to 173ºC.

  1. List the known and unknown factors. Once you're comfortable with the problem, you can write down each known and unknown variable to have a better sense of what you're working with. Be sure to pay attention to the units for mass so that what is given matches up to units in specific heat. The units will need to be both in grams or in kilograms. Here's how you identify each variable from the problem above:
  • m = 0.35 kg
  • Q = 34,700 Joules
  • ΔT = 173ºC - 22ºC = 151ºC
  • c = unknown
  1. Plug the known factors into the equation. You know the value of everything except "c", so you should plug the rest of the factors into the original equation. Since you are finding the specific heat, c, you will need to rearrange the equation and solve for "cp", Here's how you do it:
  • Rearranged equation: C = Q/mΔT
  • c = 34,700 J/(0.35 kg x 151ºC)
  1. Solve the equation. Now that you've plugged the known factors into the equation, just do simple arithmetic to solve it. The specific heat, or final answer, is 0.65657521286 J/(g x ºC).
  • c = 34,700 J/(0.35 kg x 151ºC)
  • c = 34,700 J/(52.85 kg x ºC)
  • c = 656.57521286 J/(kg x ºC)

Practice Problems

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