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Question 1

What is kinetic energy

Answer 1

Energy of motion
+ anything moving has kinetic energy

Question 2

What is potential energy

Answer 2

• Stored energy
• Dependant on condition of object

Question 3

As soon as u start moving what happens to energy ?

Answer 3

Potential energy ~> Kinetic energy

Question 4

Difference between uppercase Calorie and lowercase calorie???

Answer 4

Calorie (uppercase) = food calorie

calorie (lowercase) = thermal calorie

Question 5

What is a joule & Kilo Joule ?

Answer 5

— Derived SI unit for energy
— its a v small amount of energy, so we use Kilo Joule

Question 6

Unit conversions

1000 J = ___ KJ

Answer 6

1 KJ

Question 7

UNIT CONVERSIONS

1 J = ___ cal

Answer 7

0.2390 cal

Question 8

UNIT CONVERSIONS

4.18 J = ___ cal

Answer 8

1.00 cal (thermal) (3 sigfig)

Question 9

Unit conversions

4.18 J = ___ Cal

Answer 9

1 Cal (food) (1 sigfig)

Question 10

UNIT CONVERSIONS

1000 cal (thermal) = _____ Cal

Answer 10

= 1 Cal (food)

Question 11

What is specific heat capacity?

Answer 11

Amount of energy (heat) needed to raise 1 gram of the substance by 1 degree Celsius

Question 12

Formula for specific heat capacity ?

Answer 12

Heat capacity (C) = heat absorbed Q/ incr^ in temp T

— T can be (final T - initial T too)

Question 13

Define Open system and give example

Answer 13

Can exchange BOTH ENERGY & MATTER w surroundings

Ex. An uncovered pot of potatoes boiling on the stove
=>The system absorbs energy from the stove burner. The system loses both energy and matter when the water evaporates out of the pan in the form of steam.

Question 14

Define Closed system and give example

Answer 14

CAN exchange ENERGY NOT MATTER w/ surroundings

Ex. The pressure cooker with potatoes boiling on the stove
=> because pressure cookers are sealed. The sealed lid prevents the loss of mass, or water in the form of steam, but heat can still enter the system through contact of the pot bottom with the stove.

Question 15

Isolated system

Answer 15

CANNOT exchange ENERGY AND MATTER w/ surroundings

Ex. A hot pan of food inside an insulated box
- no heat can go in or out
- no matter can leave the box either

Ex. the pot of potatoes inside an insulated container
=>The insulation prevents the exchange of any energy or matter between the system and its surroundings.

Question 16

Universe =

Answer 16

System + surroundings

Question 17

Define system

Answer 17

Set of reactants + products being observed

Question 18

Define surroundings

Answer 18

Everything else

Question 19

What happens when a system and its surrounding interact

Answer 19

Energy and matter get exchanged

Question 20

Law of conservation of energy

Answer 20

Energy cannot be created or destroyed

it can only change form

Question 21

Good job!!

Question 22

Explain this:

Δ Etotal = Δ Esys + Δ Esurr = 0

Answer 22

total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Question 23

Explain what each letter stands for and what we use this formula for

ΔQ = m . c . ΔT

Answer 23

Q = heat ethalpy (+ absorbed by the system / - left from system basically released)

M = mass

C= specific heat capacity

ΔT = change in temp
We find by (T final - T initial)

We use this to calculate amt of heat absorbed or released

Question 24

Solve

A 1.0 g sample of copper is heated from 25.0°C to 31.0°C. How much heat did the sample absorb? The specific heat capacity of copper is 0.385 J/g°C.

Answer 24

Given:

M = 1.0 g
C = 0.385/g°C
Δt = final - initial
31.0 - 25.0 = 6.00

Q = m.c.Δt
=1.0x0.385x6.00
=>2.3 J/g°C

Question 25

Open closed or isolated The universe

Answer 25

Isolated

Question 26

Open closed or isolated Ceramic mug of hot coca

Answer 26

Open

Question 27

Open closed or isolated Sealed metal canister

Answer 27

Closed

Question 28

Open closed or isolated An auto mobile with all doors, windows and vents closed Explain why as well here

Answer 28

Closed, not isolated as it can still exchange energy not matter (ie; absorb heat) from environment

Question 29

Open closed or isolated A sealed insulated canister

Answer 29

Isolated

Question 30

What does the first law of thermodynamics mean?

Answer 30

It means if a system undergoes a change that results in an increase or decrease in the amount of energy it has, That energy must have (come from or gone to)the surroundings

Question 31

Explain ΔEuniverse = 0

Answer 31

- first law of thermodynamics states that the total energy of the universe is constant. - Energy is neither created nor destroyed; it is transformed from one type of energy to another or transferred from one object to another. So, change in the energy of the universe is zero. (Δ = change)

Question 32

So, explain step by step Why is ΔEsys = -ΔEsurr

Answer 32

Since, when defining a system, the surroundings consist of everything else in the universe you could also say that the system plus the surroundings equals the universe — universe = system + surroundings the total energy of the universe is equal to the total energy of the system plus the total energy of the surroundings: Euniverse = Esystem +Esurroundings Similarly, any change in the total energy of the universe must be equal to the change in the total energy of the system + the change in the total energy of the surroundings. Since the change in the total energy of the universe is equal to zero, the following applies: ΔEuniverse = ΔEsystem + ΔEsurroundings = 0 By re-arranging this equation, any change in the total energy of the system must be equal and opposite to the change in the total energy of the surroundings: ΔEsystem = −ΔEsurroundings

Question 33

Generally explain this ΔEsys = -ΔEsurr

Answer 33

In terms of practical situations such as warming soup on a stove, thawing ice, and any laboratory situation involving physical and chemical changes, the implications of the first law of thermodynamics are relatively straightforward: — for any system that gains energy, -> the energy must come from the surroundings — if any system loses energy, ->that energy must enter the surroundings.

Question 34

Solve : A 25.00 g sample of water is heated to a temperature of 38.0°C, absorbing 2.96kJ of heat. What was the initial temperature of the water? The specific heat capacity of liquid water is 4.19 J/g°C.

Answer 34

ALWAYS NOTICE UNITS Given : (Change the units to match, turn them both into J) Q= 2.96 kJ —> 2980 J M = 25.00 J C= 4.19 J Rearrange formula To find time — Q = mcΔT -> ΔT = Q / m x c Sub in 2960 / 25 x 4.19 = ΔT 28.26 = ΔT ΔT = 28.26 Take the final temperature we are already given and subtract the change in temperature ΔT from it 38.0 - 28.26 = 9.7 initial temperature is 9.7°C

Question 35

What is the first law of thermodynamics?

Answer 35

Energy can change form but can never be destroyed.

Question 36

What is the second law of thermodynamics ?

Answer 36

When 2 objects are in thermal contact, heat is transferred from the object at a higher temp to the object at a lower temp until they are both the same temperature

Question 37

What is thermal equilibrium

Answer 37

When both objects get to the same temp.

Question 38

Why are the categories of enthalpy changes