endo or exothermic sign of q (+ or -) melting of ice endo+ evaporationendo+ condensationexo-...

Endo or Endo or ExothermicExothermic

Sign of q Sign of q

(+ or -)(+ or -)

Melting of IceMelting of Ice endoendo ++

Evaporation Evaporation endoendo ++

condensationcondensation exoexo --

subliminationsublimination endoendo ++

(l) (l) (s) (s) exoexo --

s s l l

l l g g

S S g g

l l s s

g g l l

g g s s

Melting/fusion endo

vaporization endo

sublimination endo

freezing exo

condensation exo

deposition exo

SOLID

Freezing

Melting

condensation

vaporization

LIQUID

GAS

FUSION

The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

Heat capacityHeat capacity = the amount of energy required to raise the = the amount of energy required to raise the temperature of an object (by one degree).temperature of an object (by one degree).

Molar heat capacityMolar heat capacity = heat capacity of 1 mol of a substance = heat capacity of 1 mol of a substance..

Heat of fusion: The amount of energy released/required at the solid/liquid phase change.

Heat of vaporization: The amount of energy released/requiredfor liquid/ gas phase change.

Heat is the total amount of energy possessed by the molecules in a piece of matter. This energy is both kinetic energy and potential energy.

Temperature is proportional to the average kinetic energies of the molecules

moves spontaneously from moves spontaneously from matter with higher T to matter with matter with higher T to matter with lower Tlower T

What happens when two objects of different What happens when two objects of different temperatures come into contact?temperatures come into contact?

Which of the following has the greatest heat capacity?

100 g water or 1000 g water

Which of the following has the greatest specific heat?

100 g water or 100 g copper

1. How much heat is required to raise the temp of 205 g of waterfrom 15.2 C to 16.2 C ?

What info do we have?

m = 205 g c= 4.184 J/g C t = 16.2 – 15.2 = 1

What are we looking for ? q

What is our equation ? q = mct

q = (205 g) (4.184 J/g C) (1 C)

858 J

2. Calculate the amount of heat released when 25 g of waterAt 25 C is cooled to 0 C ?

What info do we have?

m = 25 g c= 4.184 J/g C t = 25 – 0 = 25 C

What are we looking for ? q

What is our equation ? q = mct

q = (25 g) (4.18 J/g C) (25 C)

2615 J

3. What mass of 67.5 C iron must be added to 235 g of 5.00 C waterTo make the final temp of both come out to be 15 C ?

What do we have?

Iron c = .444 J/g C m = ? gIron initial T = 67.5 CIron final T = 15 C

t = 67.5 – 15 = 52.5

Water c = 4.184 J/g CWater m = 235 gWater initial T = 5 CWater final T = 15 C

t= 15 – 5 = 10 C

Heat lost = Heat gained q = q

mct= mct

(?g) (.444g) (52.5) = (235g)(4.184 J/g C)(10 C)

mass = (235g)(4.184 J/g C)(10) (.444J/g C)(52.5 C)

421.8 g Fe

4. A 195 g aluminum engine part at an initial temperature of 3.00 C absorbs 40 KJ of heat. What is the final temperature of the part?

What info do we have?

m = 195 g c= .897 J/g C T initial= 3.00 C

What are we looking for ? T final

What is our equation ? q = mct

40,000J = (195 g) (.897 J/g C) (Tfinal – 3.00 C)

q = 40 KJ 40,000J

q= mc(Tfinal-Tinit)

40000J = Tfinal – 3.00 C(195g)(.897J/g C)

40000J + 3.00 C = Tfinal (195g)(.897J/g C)

5. When 300 J of energy is lost from 125 g object, the temperature decreases from 45 C to 40 C. What is the specific heat of this object?

What info do we have?

m = 125 g T final = 40C T initial= 45.00 C

What are we looking for ? c specific heat

What is our equation ? q = mct

300J = (125 g) (c) (5 C)

q = 300 J

300J = c(125g)( 5 C)

.48J/g C

6. The specific heat of lead (Pb) is 0.129 J/g C. Find the amount ofheat released when 2.4g of lead are cooled from 37.2 C to 22.5 C?

What info do we have?

m = 2.4 g T final = 22.5C T initial= 37.2 C

What are we looking for ? q

What is our equation ? q = mct

q = (2.4 g) (.129 J/g C) (22.5C – 37.2 C)

c = .129J/g C

4.6 J

7. How many kJ of energy are needed to raise the temperatureof 165 g water from 10.5 C to 47.32 C?

What info do we have?

m = 165 g T final = 47.32 C T initial= 10.5 C

What are we looking for ? q in units of kJ

What is our equation ? q = mct

q = (165 g) (4.184 J/g C) (47.32 C – 10.5 C) =

c = 4.18J/g C

25395 J 25395 J x 1 kJ = 25.4 KJ 1000 J

8. How much heat is absorbed by 15 g of ice being melted?

H2O (s) H2O (l) Hf = 6.01 kJ/1 mol

15 g x 1 mol = .833 mol18 g

.833 mol x 6.01 kJ/mol = 5 kJ

9. How much heat is necessary to change 5.0 g of water at 100 C to to steam at 100 C?

H2O (l) H2O (g) Hv = 2260 J/g

5g x 2260 J/g = 11300 J

Calculate Molar Mass of

Calcium Phosphate

Ammonium Sulfate

CaPO4

(40) + (31) + 4(16) = 135 g/mol

(NH4)2SO4

(18)2 + 32+ 4(16) = 132 g/mol

How many moles of Na2CO3 are thee in 10L of 2.0M solution

2.0 mol = X mol1 liter 10 liter

= 20 moles

Find the molarity of a solution containing 59 g HClIn 500 ml of water.

First convert grams to moles

59 g HCl x 1 mole HCL36 g

= 1.6 mol

Molarity is moles per liter

1.6 mole x 1000 ml500 ml 1 liter

= 3.2M

What volume (in ml) of 12.o M HCl is needed to contain 3.00 moles HCl

12.0 mol = 3.0 mol1 liter X liter

X = 3.012

X = .25 liter

X = 250ml

How many moles of gas would be present in a gas Trapped within a 100 ml vessel at 25 C at a pressure Of 2.50 atm

PV = nRT n = PVRT

2.5atm (0.1L)(0.821)(298K)

What volume will 1.27 moles of helium gasoccupy at STP

1.27 mol x 22.4 L mol

28.5 L