1. 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

2. 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

3. 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

4. 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

5. 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

6. 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 = 5 g
c = water = 4.18 j/goC
∆t = 75 – 25 = 50 oC
Q = (5g) (4.18 j/goC)(50 oC)

7. 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?

8. q = mc∆t q = 48 J m = 12 g c = ? ∆t = 40 – 20 = 20 oC (12 g) (20 oC)
plug in heat equation: J = 12 g (c) 20 oC
rearrange to solve for c
48 J
___________ = c
(12 g) (20 oC)

9. 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?

10. -------------------------------
Answer: 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?

11. Set up: q= (4 g)(.19 J/g oC)(29 oC)
Answer: 22J
Set up: q= (4 g)(.19 J/g oC)(29 oC)

12. 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

13. 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 = ∆t
(10g) (4.18 j/g oC)
47.8 oC = ∆t ∆t = (final temp – initial temp)
47.8 = final temp – 25 oC
= final temp

14. 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!