Heat Capacity Example Problem

Heat capacity is the amount of heat energy required to change the temperature of a substance. This example problem demonstrates how to calculate heat capacity.

Problem: Heat Capacity of Water From Freezing to Boiling Point

What is the heat in joules required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C? What is the heat in calories?

Useful information: specific heat of water = 4.18 J/g·°C
Solution:

Part I

Use the formula

q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
q = (25 g)x(4.18 J/g·°C)[(100 C - 0 C)]
q = (25 g)x(4.18 J/g·°C)x(100 C)
q = 10450 J
Part II
4.18 J = 1 calorie
x calories = 10450 J x (1 cal/4.18 J)
x calories = 10450/4.18 calories
x calories = 2500 calories
Answer:
10450 J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C.

Tips for Success

• The most common mistake people make with this calculation is using incorrect units. Make certain temperatures are in Celsius. Convert kilograms to grams.
• Be mindful of significant figures, particularly when working problems for homework or an exam.