Specific Heat Example Problem


Molten Glass

The energy required to heat something is proportional to the mass and temperature change of the material. The proportionality constant is called specific heat.
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Specific heat is the amount of heat per unit mass needed to increase the temperature of a material by one degree Celsius or Kelvin. These three specific heat example problems will show how to find the specific heat of a material or other information involving the specific heat.

Specific Heat Equation

The equation most commonly associated with specific heat is

Q = mcΔT

where
Q = Heat energy
m = mass
c = specific heat
ΔT = change in temperature = (Tfinal – Tinitial)

A good way to remember this formula is Q = “em cat”

Basically, this equation is used to determine the amount of heat added to a material to raise the temperature some amount (or the amount lost as the material cools).

This equation only applies to materials that stay in the same state of matter (solid, liquid, or gas) as the temperature changes. Phase changes require additional energy considerations.

Specific Heat Example Problem – Find the Amount of Heat

Question: A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

Solution: First, let’s the variables we know.

m = 500 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C

Plug these values into the specific heat equation from above.

Q = mcΔT

Q = (500 grams)·(0.129 J/g°C)·(50 °C)

Q = 3225 J

Answer: It took 3225 Joules of energy to heat the lead cube from 25 °C to 75 °C.

Specific Heat Example Problem – Find the Specific Heat

Question: A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?

Solution: List the information we know.

m = 25 grams
ΔT = 200 °C
Q = 2330 J

Place these into the specific heat equation.

Q = mcΔT

2330 J = (25 g)c(200 °C)

2330 J = (5000 g°C)c

Divide both sides by 5000 g°C

specific heat example math step 1

c = 0.466 J/g°C

Answer: The specific heat of the metal is 0.466 J/g°C.

Specific Heat Example Problem – Find the Initial Temperature

Question: A hot 1 kg chunk of copper is allowed to cool to 100°C. If the copper gave off 231 kJ of energy, what was the initial temperature of the copper? The specific heat of copper is 0.385 J/g°C.

Solution: List our given variables:

m = 1 kg
Tfinal = 100 °C
Q = -231 kJ (The negative sign is because the copper is cooling and losing energy.)
c = 0.385 J/g°C

We need to make our units consistent with the specific heat units, so let’s convert the mass and energy units.

m = 1 kg = 1000 grams

1 kJ = 1000 J
Q = -231 kJ · (1000 J/kJ) = -231000 J

Plug these values into the specific heat formula.

Q = mcΔT

-231000 J = 1000 g · (0.385 J/g°C) · ΔT

-231000 J = 385 J/°C · ΔT

Specific Heat Example Problem Math Step 2

ΔT = -600 °C

ΔT = (Tfinal – Tinitial)

Plug in the values for ΔT and Tfinal.

-600 °C = (100 °C – Tinitial)

Subtract 100 °C from both sides of the equation.

-600 °C – 100 °C =  – Tinitial

-700 °C = – Tinitial

Tinitial = 700 °C

Answer: The initial temperature of the copper chunk was 700 °C.

 

 

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