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 °C to 100 °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 °C to 100 °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.