# Chapter 11

• positive q = system has gained heat from the surroundings (endothermic)
• negative q = system has released heat to the surroundings (exothermic)
q depends on the change within the system, no on how the change occurs

• specific heat = measure the change in temperature that a known mass undergoes as it loses or gains a specific quantity of heat
Specific heat of water is 4.184 J/gºC

Formulas to know:

```                         q  =  m .  C  . ∆T

and

qlost  =  qgained

which means,

mlost  .  Clost  .  ∆Tlost  =    mgained  .  Cgained  .  ∆Tgained  ```

∆T means a change in temperature represented by taking the final temperature minus the initial temperature.

Example: The temperature of a piece of copper with a mass of 95.4 grams increases from 25.0ºC to 48.0ºC when the metal absorbs 849 J of heat. What is the specific heat of copper?

Variables:
Mass = 95.4 grams
Heat = 849 J
Initial Temperature = 25.0ºC
Final Temperature = 48.0ºC
Specific Heat (C) = ?

q = m . C . ∆T

849 J = (95.4 grams) (C) (48.0ºC - 25.0ºC)

849 J = (95.4 grams) (C) (23ºC)

849 J = 2194.2 C

849 J/2194.2 = 2194.2/2194.2 C

C = 0.387 J/gºC

Example:Calculate the final temperature if the following mixtures of water are placed together in the same cup: Cup A contains 25 grams of water at 23ºC and Cup B contains 50 grams of water at 89ºC.

qlost = qgained

mlost . Clost . ∆Tlost = mgained . Cgained . ∆Tgained

(50 grams) (4.184 J/gºC) (89ºC - x) = (25 grams) (4.184 J/gºC) (x - 23ºC)

(209.2) (89ºC - x) = (104.6) (x - 23ºC)

18618.8 - 209.2x = 104.6x - 2405.8

18618.8 + 2405.8 = 104.6x + 209.2x

21024.6 = 313.8x

21024.6/313.8 = 313.8/313.8 x

x = 67ºC

Calculate the amount of heat is required to raise the temperature of 50 grams of ice at -10ºC to steam at 120ºC.

Heat of fusion = 334 J/g
Heat of vaporization = 2260 J/g
specific heat of water = 4.184 J/gºC
specific heat of ice = 2.1 J/gºC
specific heat of steam = 1.7 J/gºC
1. Must raise the temperature to ice at 0ºC
2. Change the state from solid to liquid.
3. Raise the temperature from liquid at 0ºC to 100ºC.
4. Change the state from a liquid to a gas.
5. Raise the temperature from gas at 100ºC to 120ºC.

Raise the temperature → q = m•C•∆T

Change the state → Energy transferred = m•Hf or v

1. q = (50 g)(2.1 J/gºC)(0ºC - -10ºC) = 1,050 J
2. e.t. = (50 g)(334 J/g) = 16,700 J
3. q = (50 g)(4.184 J/gºC)(100ºC - 0ºC) = 20,920 J
4. e.t. = (50 g)(2260 J/g) = 113,000 J
5. q = (50 g)(1.7 J/gºC)(120ºC - 100ºC) = 1,700 J

Total energy = the sum of all the energy = 153,370 J

Hess's Law

The idea behind Hess's law is illustrated as follows:

Reaction equation (1) with ∆H1
+ Reaction equation (2) with ∆H2
is Reaction equation (3) with ∆H3 = ∆H1 + ∆H2