Specific Heat Example Problem

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Helmenstine, Todd. "Specific Heat Example Problem." ThoughtCo, Apr. 12, 2017, thoughtco.com/specific-heat-example-problem-609531. Helmenstine, Todd. (2017, April 12). Specific Heat Example Problem. Retrieved from https://www.thoughtco.com/specific-heat-example-problem-609531 Helmenstine, Todd. "Specific Heat Example Problem." ThoughtCo. https://www.thoughtco.com/specific-heat-example-problem-609531 (accessed October 19, 2017).
Heating copper
Specific heat can be calculated if you know how much energy it takes to change temperature. Opla / Getty Images

This worked example problem demonstrates how to calculate the specific heat of a substance when given the amount of energy used to change the substance's temperature.

Specific Heat Equation and Definition

First, let's review what specific heat is and what equation you use to find it. Specific heat is defined as the amount of heat per unit mass needed to increase the temperature by one degree Celsius (or by 1 Kelvin).

Usually, the lowercase letter "c" is used to denote specific heat. The equation is written:

Q = mcΔT (remember by thinking "em-cat")

where Q is the heat that is added, c is specific heat, m is mass and ΔT is the change in temperature. The usual units used for quantities in this equation are degrees Celsius for temperature (sometimes Kelvin), grams for mass, and specific heat reported in calorie/gram °C, joule/gram °C, or joule/gram K. You can also think of specific heat as heat capacity per mass basis of a material.

When working a problem, you'll either be given the specific heat values and asked to find one of the other values or else asked to find specific heat.

There are published tables of molar specific heats of many materials. Note the specific heat equation does not apply for phase changes. This is because the temperature does not change.

Specific Heat Problem

It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C.

What is the specific heat in Joules/g·°C?

Solution:
Use the formula

q = mcΔT

where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

Putting the numbers into the equation yields:

487.5 J = (25 g)c(75 °C - 25 °C)
487.5 J = (25 g)c(50 °C)

Solve for c:

c = 487.5 J/(25g)(50 °C)
c = 0.39 J/g·°C

Answer:
The specific heat of copper is 0.39 J/g·°C.