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:

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.