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

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

specific heat of water is 40184j/g celsious

4.184

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

100416

what is that

that is the answer i got when i

(800.0)(4.184)(30.0)

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

how am i adding in kilojoules/?

don't i need to convert to kilojoules??

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

the answer is 3rd one

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

yaaa dude

heat difference is the delta t right?

yaaaa delta T =T2-T1

its temperature difference not heat difference

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

i learn best by return demonstration

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

find out S all other have given in this

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

yaaa right

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

1000?

what 1000? mean

is that my heat?

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

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

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

LOL!!!!

are you going to be online for a bit?

why???

u have another question?

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

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

a

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

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

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

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

so..solving for mass.

yaaaa solve it

m=q/cpxdelta t

yaaa

is this my cp 4.50 x 103 J??

no wait..that's my heat.

good

good to what ? is it cp or heat?

heat only as u said

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

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

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

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

yaaa89.6 g

yay!! ok..

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

enough man

write balanced equation

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

pellet is a type-o

i didnt get ur question properly chlorine or chloride

ok may be its correct

chlorine + sodium =sodium chloride

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

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

u solve it dude

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

k

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

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

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

tahnk you for your help

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