1. Unit 04 - Heat
Ht 1 03 – Energy Changes in one substance & Energy transfer using conduction

2. Calculating Energy Changes in One Substance:
When energy is given off in the form of heat, this value can be measured or calculated
Recall that the base unit for energy is the joule (J)
If the amount of heat given off by a substance is to be measured, its specific heat capacity must also be known.
Specific heat capacity is defined as the amount of heat required to raise 1 g of a substance 1˚C. Therefore the units for a substance’s specific heat capacity is J/g˚C.
The specific heats of several substances will be provided for you (in your orange data table booklet)

3. Calculating Energy Changes:
There is a WAY COOL FORMULA we can use to calculate the amount of energy absorbed or released from a substance when the temperature changes. It is:
Q = mc∆t
Q = energy (J) absorbed or released
m = mass of the substance (g)
c = specific heat capacity (J/g°C)
∆t = change in temperature (°C)
tfinal - tinitial

4. Calculating Energy Changes:
Please note:
When using the Q = mc∆t formula:
If heat is absorbed, Q will be positive.
If heat is released, Q will be negative.
The value of Q itself will always be positive. You need to understand mentally if Q is representing heat being absorbed or heat energy being released; the negative sign is simply representing if it’s absorbed or released

5. Example 1:
What quantity of heat is required to raise the temperature of 100.g of liquid water 10.°C?

6. Example 2:
What quantity of heat is released when 420.0g of lead cools from 85°C to 25°C? The specific heat capacity of lead is 0.13J/g°C.

7. CALORIMETRY – Calculating heat transferred from one object to another
Remember this?
When heat is transferred from one substance to another, it can be calculated  this process is called calorimetry

8. CALORIMETRY – Calculating heat transferred from one object to another
The formula for calculating calorimetry questions is as follows:
Qsystem = - Qsurroundings
Or Q1 = - Q2
Remember:
Q = mc∆t
Therefore:
mc∆tsystem = - mc∆tsurroundings Or
mc∆t1 = - mc∆t2

9. Example 3
How much heat is absorbed by a 25.0-g sample of gold at 25.0 °C when it is immersed in boiling water? The specific heat capacity of gold is J/ g·°C and the specific heat capacity of water is J/ g·°C.
Ans: kJ

10. Example 4
A piece of stainless steel of mass 25.0 g at 88.0 °C was placed in a calorimeter that contained g of water at 20.0 °C. If the temperature of the water rose to 21.4 °C, what is the specific heat capacity of the stainless steel? The specific heat capacity of water is J/ g·°C
Ans: J/ g ·°C

11. Example 5
What is the mass of a piece of graphite (c = 0.71 J/g ·°C) at 92.0 °C if it was placed in a calorimeter that contained g of water at 12.0 °C. If the temperature of the water rose to 16.8 °C? The specific heat capacity of water is J/ g·°C
Ans: g

12. Let’s Try A Toughie – Example 6
A 65.0 g piece of iron (ciron = J/g·oC) at 525° C is put into 635 grams of water at 15.0° C.
What is the final temperature of the water and the iron?
Tf = 20.5° C