- 184 Views
- Uploaded on
- Presentation posted in: General

Specific Heat

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Different substances have different abilities to store energy. They are said to have different heat capacities.

- Heat capacity is defined as the amount of heat required to change the temperature of the substance by one degree Celsius (or Kelvin).

Most metals have low specific heats, while nonmetal compounds & mixtures such as water, wood, soil, & air have relatively high specific heats.

- A device that measures temperature changes in surroundings
- Heat transferred by physical and chemical changes can be measured using a process called calorimetry.

- Calorimetry: Process for determining the amount of heat energy released or absorbed in a chemical or physical change.
- Enthalpy : Change in Heat energy Symbolized by H
- Entropy : extensive property which measures the degree of disorder.

- calorie - the amount of heat required to change the temperature of 1 gram of pure liquid water by one degree Celsius.
- Food calorie (Calorie, big calorie) - is equal to 1000 calories or one kilocalorie
- Joule - International System unit of energy. There are 4.18 Joules in one calorie.
- Kilojoule - 1000 joules
- For heat conversions use:
- 1Kcalorie= 1Calorie = 1000 calories
- 1calorie = 4.18J

Foods and Heat Energy

- 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.
- Energy in our bodies comes from carbohydrates and fats (mostly).
- Intestines: carbohydrates converted into glucose:
- C6H12O6 + 6O2 6CO2 + 6H2O, DH = -2816 kJ
- Fats break down as follows:
- 2C57H110O6 + 163O2 114CO2 + 110H2O, DH = -75,520 kJ
- Fats contain more energy; are not water soluble, so are good for energy storage.

Problem #1

- 1Kcalorie= 1Calorie = 1000 calories
- 1calorie = 4.18J

1 1Cal 6.00x104cal

Problem #2

- 1Kcalorie= 1Calorie = 1000 calories
- 1calorie = 4.18J

1 1Cal 1cal

Since the calorie is defined in terms of water, the heat capacity for liquid water is 1 cal/g oC.

- This also equates to 4.18 J/g oC.

The greater the mass of the object, the greater its heat capacity.

A massive steel cable on a bridge requires much more heat to raise its temperature 1ºC than a small steel nail does.

Different substances with the samemass may have different heat capacities.

On a sunny day, a 20-kg puddle of water may be cool, while a nearby 20-kg iron sewer cover may be too hot to touch.

When a hamburger is burned in a calorimeter, 2000. g of water increases in temperature by 30.0 oC. How many Calories are in the hamburger?

Q=(m) (Cp) (T)

- Cwater = 1 cal/g oC or 4.18 J/g oC
- T = 30 C

- Q = 2000. g x 1 cal/g oC x 30.0 oC = 60000cal
- 60000cal x 1Cal = 60.0 Cal
1 1000cal

The temperature of 2500 grams of mercury rises from 20 oC to 60oC when it absorbs 13,794 joules of heat. Calculate the specific heat capacity of the mercury.

Q=(m) (Cp) (T)

- Cmercury =?
- ΔT = 40 oC

- 13,794 J = 2500g x (Cp) x 40.0 oC
- 13,794 J = 100000g oC x (Cmercury)
- (Cmercury)=0.138 J/g oC

- An 800-gram block of lead is heated in boiling water (100 oC) until its temperature is the same as the boiling water. The lead is then removed from the boiling water and dropped into 250 grams of cool water at 12.2 oC. After a short time, the temperatures of both lead and water levels out at 20.0 oC.
- Calculate the amount of heat (in Joules) gained by the cool water.
- Q=(m) (Cp) (T)

m= 250g

Ti= 12.2 C and Tf= 20.0 C

T = 7.8 C

Cwater= 4.18 J/gC

Q = (250g) ( 4.18 J/gC)( 7.8 C )

Q =8151J

- An 800-gram block of lead is heated in boiling water (100 oC) until its temperature is the same as the boiling water. The lead is then removed from the boiling water and dropped into 250 grams of cool water at 12.2 oC. After a short time, the temperatures of both lead and water levels out at 20.0 oC.
Calculate the specific heat capacity of the lead based on these measurements, assuming that no heat was lost in the process.

- Q=(m) (Cp) (T)

m= 800g

T = 80 C

CPb= ?

Q gained by water= Q lost by Pb = 8151J

8151J = (800g) (CPb) (80 C)

CPb = 0.127 J/gC

At the freezing or boiling point two phases of matter can exist at the same temperature

To make the change from one phase to another more energy will be absorbed (boiling or melting) or lost (condensing or freezing) without a change in temperature

This is because this energy is used merely to overcome the bonds of one state and move to the new state creating a change in potential energy.

Heat of vaporization

Gas/vapor

D PE

Boiling/vaporization

D KE

condensation

Heat of fusion

D PE

liquid

melting

D KE

D KE

Solid

freezing

100oC

0oC

Time

Instead of specific heat (C) we use enthalpy (∆H) for calculating heat during phase changes.

Heat of fusion/solidification is the heat required to move from solid liquid

Hfus = –Hsolid

Heat of vaporization/condensation is the heat required to move from liquid gas

Hvap = –Hcon

Q = m x H

- Q = amount of heat energy (joules or calories)
- m = mass of substance (grams)
- H = enthalpy of fusion (Hf) or vaporization (Hv)

- constants:
Specific heat of ice = 2.09 J/g·ºC

Specific heat of water = 4.18 J/g·ºC

Specific heat of steam = 2.03 J/g·ºC

Heat of fusion of water = 334 J/g

Heat of vaporization of water = 540 J/g

- Calculate the mass of ice (in grams) that will melt at 0ºC if 2.25 kJ of heat are added. (Hf= 334 J/g)
Q = m x Hf

- Q = 2250 J
- Hf = 334 J/g
- m = Q /Hf
- m = 2250 J / 334 J/g

m = 6.74 g

- Calculate the mass of water vapor (in grams) at 100ºC that can be condensed into liquid at 100ºC if 55.0 kJ of heat is removed. (Hv = 2257 J/g)
Q = m x Hv

- Q = 55000 J
- Hv= 2257 J/g m
- m = Q /Hv
- m = 55000 J / 2257 J/g
- m = 24.4 g

- How much heat does it take to turn a 20 g chunk of ice at -40oC into 20 g of steam at 120oC?
- This is a 5 step problem, each segment of the graph must be calculated separately and then added together to get a total heat absorbed.

-40oC

- How much heat does it take to turn a 20 g chunk of ice at -40oC into 20 g of ice at 0oC?
Q=(m) (Cp) (T)

m= 20g

Ti= -40 C and Tf= 0 C

T = Tf - Ti =0 C – (-40.0 C)= 40 C

Cice= 2.09 J/g·ºC

Q = (20g) ( 2.09 J/gC)( 40 C )

Q = 1672J (+Q indicates heat gained or endothermic)

- How much heat does it take to turn a 20 g chunk of ice at 0oC into 20 g of water at 0oC?
Q=(m) (Hf)

m= 20g

Hf= 334 J/g

Q = (20g) ( 334J/g)

Q = 6680J (+Q indicates heat gained or endothermic)

- How much heat does it take to turn a 20 g water at 0oC into 20 g of water at 100oC?
Q=(m) (Cp) (T)

m= 20g

Ti= 0 C and Tf= 100 C

T = Tf - Ti = 100 C – 0C= 100 C

Cwater= 4.18 J/g·ºC

Q = (20g) ( 4.18 J/gC)( 100 C )

Q = 8360J (+Q indicates heat gained or endothermic)

- How much heat does it take to turn a 20 g water at 100oC into 20 g of water vapor at 100oC?
Q=(m) (Hv)

m= 20g

Hf= 334 J/g

Q = (20g) ( 334J/g)

Q = 6680J (+Q indicates heat gained or endothermic)

- Entropy is a measure of how chaotic a system is. The lessorder that is present the more entropy. In terms of states of matter:

Liquid

Gas

Solid

Low Entropy

High Entropy