Specific Heat
- 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