HEAT EQUATION
(in Table T)
Q = mc∆t
Q = heat energy in JOULES (J)
m = mass of the sample in GRAMS (g)
C = specific heat in J/goC
∆t = change in temperature (oC)
or
final temp – initial temp
Using the Heat Equation
How many Joules of Heat are needed to
raise the temperature of 25 g of water from
10 oC to 60 oC?
Step 1: Write the heat equation
q = mc∆t
Using the Heat Equation
How many Joules of Heat are needed to raise the
temperature of 25 g of water from 10 oC to 60 oC?
q = mc∆t
Step 2: write the information given
q=?
m = 25g
c = water = 4.18 J/goC (Table B)
∆t = final temp – initial temp
= 60 oC – 10 oC
= 50 oC
How many Joules of Heat are needed to raise the
temperature of 25 g of water from 10 oC to 60 oC?
q = mc∆t
q=?
m = 25g
c = water = 4.18 J/goC (Table B)
∆t = final temp – initial temp
= 60 oC – 10 oC
= 50 oC
Step 3. Substitute values into the equation and solve for
the unknown
q = (25 g) ( 4.18 J/goC)( 50 oC)
q = 5225 J
= ______ KJ
Practice using the heat equation
How many Joules of heat are given off when
5 grams of water cools from 75 to 25 oC?
q = mc∆t
Practice using the heat equation
How many Joules of heat are given off when
5 grams of water cools from 75 to 25 oC?
q=?
m=5g
c = water = 4.18 j/goC
∆t = 75 – 25 = 50 oC
Q = (5g) (4.18 j/goC)(50 oC)
Answer: q = 1045 J
Try another!
12 grams of unknown material are heated
from 20 to 40 oC and absorb 48 Joules of
heat. What is the specific heat (c) of the
unknown material?
q = mc∆t
q = 48 J
m = 12 g
∆t = 40 – 20 = 20 oC
plug in heat equation:
rearrange to solve for c
c = ?
48 J = 12 g (c) 20 oC
48 J
___________
(12 g) (20 oC)
=
c
Answer: 0.2 J/g0C = c
Try another:
4 grams of glass are heated from 0 to 42
degrees Celsius and absorb 32 Joules of
heat. What is the specific heat of glass?
Answer: 0.19 J/g0C = c
Set up: 32J / (4 g)(42 oC) = c
------------------------------Try another:
If the 4 g of glass from the last problem are
heated from 41 to 70 degrees Celsius,
how much heat will it absorb?
Answer: 22J
Set up: q= (4 g)(.19 J/g oC)(29 oC)
Try a trickier one….
10 grams of water at an initial temperature of 25
oC is heated and absorbs 2000J of heat. What
is its final temperature?
Hint: ∆t = (final temp – initial temp)
–
–
–
use ∆t as your unknown value
first solve for ∆t using the heat equation
then knowing ∆t and initial temp
(given), solve for final temp
Answer: final temp = 72.8 oC
Set up:
q = 2000J
m = 10 g c = 4.18 J/g oC ∆t = ?
2000J = (10g)(4.18 j/g oC) ∆t
2000J
(10g) (4.18 j/g oC)
=
∆t
47.8 oC = ∆t ∆t = (final temp – initial temp)
47.8 = final temp – 25 oC
47.8 + 25 = final temp
For Homework:
100 g of water cools to a final temperature of
50 degrees Celsius, releasing 4.2
kilojoules* of heat. What was the initial
temperature of the water?
* Convert to JOULES first!