Chapter 11

q depends on the change within the system, no on how the change occurs



Specific heat of water is 4.184 J/gC

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 25.0C to 48.0C 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 = 25.0C
Final Temperature = 48.0C
Specific Heat (C) = ?

q = m . C . ∆T

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

849 J = (95.4 grams) (C) (23C)

849 J = 2194.2 C

849 J/2194.2 = 2194.2/2194.2 C

C = 0.387 J/gC




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 23C and Cup B contains 50 grams of water at 89C.

qlost = qgained

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

(50 grams) (4.184 J/gC) (89C - x) = (25 grams) (4.184 J/gC) (x - 23C)

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

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 = 67C



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


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

Raise the temperature → q = mC∆T

Change the state → Energy transferred = mHf or v

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