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Specific Heat Example Problem

Worked Example Problems

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Specific heat can be calculated if you know how much energy it takes to change temperature.

Specific heat can be calculated if you know how much energy it takes to change temperature.

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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.

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

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.

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