Question

How many kilojoules of energy are absorbed when 800.0g of water is heated from 20.0º C to 50.0º C?

Answer 1

specific heat of water do you know they have not mentioned !!!

Answer 2

specific heat of water is 40184j/g celsious

Answer 3

4.184

Answer 4

Q=M*S*Delta T
put all the values in this equation
m for mass S for specific heat of water and Delta T is difference in temperature

Answer 5

100416

Answer 6

what is that

Answer 7

that is the answer i got when i

Answer 8

(800.0)(4.184)(30.0)

Answer 9

How many kilojoules of energy are absorbed when 800.0g of water is heated from 20.0º C to 50.0º C?
Question 3 options:

1) 39.0 kJ

2) 163 kJ

3) 100 kJ

4) 62.8 kJ

Answer 10

how am i adding in kilojoules/?

Answer 11

don't i need to convert to kilojoules??

Answer 12

yaaa the unit for 4.184 is joules
the answer you got is in joules that is 100416 J
1000J=1Kj
100416J=100416/1000=100.416KJ

Answer 13

the answer is 3rd one

Answer 14

ok..just so i understand. To convert into kj, just divide my final answer by 1000 ( which is the 1kj=1000):?

Answer 15

yaaa dude

Answer 16

heat difference is the delta t right?

Answer 17

yaaaa delta T =T2-T1

Answer 18

its temperature difference not heat difference

Answer 19

ok...mind if you watch while i walk through another question?

Answer 20

i learn best by return demonstration

Answer 21

400.0g of a metal absorbs 10000. J of heat energy and its temperature rises from 20.0º C to 103.0º C. What is the specific heat of the metal?
Question 4 options:

1) 0.301 J/gº C

2) 0.255 J/gº C

3) 3.32 J/g º C

4) 0.243 J/g ºC

Answer 22

find out S all other have given in this

Answer 23

so i'm using the specific heat formula...heat/massxdelta t

Answer 24

yaaa right

Answer 25

mass..400.0 delta t =20-103=83...which is my heat?

Answer 26

1000?

Answer 27

what 1000? mean

Answer 28

is that my heat?

Answer 29

S=q/Delta T*M =10000/83*400 =0.3J/g C

Answer 30

no its 10000J right why 1000J u want to take

Answer 31

oh yea..sorry...didn't add the extra zero..

Answer 32

LOL!!!!

Answer 33

are you going to be online for a bit?

Answer 34

why???

Answer 35

u have another question?

Answer 36

i will have another one. i found this chapter a little difficult to understand.

Answer 37

and the teacher SUCKS!!!!!! so we're having to self teach

Answer 38

a

Answer 39

right now i'm doing an open book quiz and the books isn't much help either, it really has been worse case scenario

Answer 40

and our test is tomorrow...so want to make sure i can grasp these concepts

Answer 41

i thank you for your help...:-)

Answer 42

A sample of water releases 4.50 x 103 J of heat energy and its temperature drops from 80.0ºC to 68.0ºC. What is the mass of the water?
Question 5 options:

1) 89.6g

2) 13.4g

3) 15.8g

4) 1570g

Answer 43

so..solving for mass.

Answer 44

yaaaa solve it

Answer 45

m=q/cpxdelta t

Answer 46

yaaa

Answer 47

is this my cp 4.50 x 103 J??

Answer 48

no wait..that's my heat.

Answer 49

good

Answer 50

good to what ? is it cp or heat?

Answer 51

heat only as u said

Answer 52

m=q/Sxdelta t
where S =4.184J for water
tellme answer now

Answer 53

89.627...since i have to round to sig figs..then 90g?

Answer 54

here s is called specific heat of water got it good its correct

Answer 55

wait no..that's not one of the answers..so 89.6g?

Answer 56

yaaa89.6 g

Answer 57

yay!! ok..

Answer 58

3.17g of sodium combines with chlorine to form 8.00g of sodium chloride. What is the mass of chlorine in this sample of sodium chloride?
Question 6 options:

1) 3.17g

2) 4.83g

3) 8.00g

4) 11.17g

Answer 59

enough man

Answer 60

write balanced equation

Answer 61

4.83g..pellet..that as an easy one:-P

Answer 62

pellet is a type-o

Answer 63

i didnt get ur question properly chlorine or chloride

Answer 64

ok may be its correct

Answer 65

chlorine + sodium =sodium chloride

Answer 66

i think it was asking how much chlorine was needed for sodium chloride to be 8.00

Answer 67

The specific heat of aluminum is 0.900J/gºC. A 400.0g sample of aluminum at 20.0ºC absorbs 2520. J of heat energy. What is the new temperature of the aluminum?
Question 9 options:

1) 7.00ºC

2) 13.0ºC

3) 20.0ºC

4) 27.0ºC

Answer 68

u solve it dude

Answer 69

assuming u dont know delta u solve i will tell u how u have to do next
can u find delta T

Answer 70

k

Answer 71

i'm not sure where to start..am i converting the 2520j to celsius/??

Answer 72

LOL!!!delta T=Q/M*s u do now

Answer 73

i will go dude bye do wll u know it very well

Answer 74

tahnk you for your help