Presentation is loading. Please wait.

Presentation is loading. Please wait.

Specific Heat Capacity. ANNOUNCEMENTS Unit 9 test tomorrow!

Similar presentations


Presentation on theme: "Specific Heat Capacity. ANNOUNCEMENTS Unit 9 test tomorrow!"— Presentation transcript:

1 Specific Heat Capacity

2 ANNOUNCEMENTS Unit 9 test tomorrow!

3 Take 1

4 Things heat/cool at different rates Everything changes temperature when it is heated or cooled, but certain things change temperature faster/slower than others. For example, put a metal skillet on a stove and it heats up quickly, so fast that it can burn you. Put water in a skillet on the stove, and the metal skillet heats faster than the water.

5 Why? Water is more resistant to a change in temperature than metal. In other words, water has a higher heat capacity. Heat capacity- The quantity of heat energy to change the temperature of a substance by 1°C. Every different element has a specific heat capacity.

6 Some heat capacities SubstanceSpecific Heat Capacity Lead.128 J/g °C Gold.128 J/g °C Silver.235 J/g °C Copper.385 J/g °C Iron.449 J/g °C Aluminum.903 J/g °C Water4.18 J/g °C Do you notice a difference in heat capacity between water and the metals?

7 So? Water has a much higher heat capacity than the metals in the table. What does that mean? (Review of the definition of heat capacity) Water takes more energy to heat 1 °C than the metals.

8 Apply If water has a higher specific heat capacity than the metals that make up the soil and rocks around us, why might a city by the water have lower temperature than a city that is surrounded by land?

9 Solving problems with heat capacity Often in chemistry we want to know how much heat must be absorbed or lost in order for a compound to change temperatures. This is why we have an easy formula to find heat gained and heat lost.

10 Formula q= m*C*ΔT q=heat m= mass of substance C= specific heat capacity ΔT=change in temperature SIDENOTE: Δ means change in any situation

11 Problem q=mCΔT How much heat must be absorbed for 2 grams of copper to heat from 25 °C to 1084 °C? The specific heat of copper is.385 J/g °C. q= m= C= ΔT= ? 3g.385 1084-25=1059 °C

12 Solved q=m x C x ΔT ?=(3) x (.385) x (1059) = In other words, it takes 1223.14 Joules of energy to heat copper 1059 °C 1223.14

13 Guided Now lets work a few more problems on the board together. Next- Practice work to really get it down.

14 Remastery time!

15 Last minutes What did we learn? Top ten: how’d we do?


Download ppt "Specific Heat Capacity. ANNOUNCEMENTS Unit 9 test tomorrow!"

Similar presentations


Ads by Google