Question

1.0 kg of water is warmed from room temperature (25.0°C) to boiling, then half of its initial volume is vaporized. How many joules of thermal energy must the stove provide to do this? Cwater = 2.26×106 J/(kg·°C); Hv, water = 2.26×106 J/kg

Answer 1

Welcome to openstudy!

Answer 2

Heat needed to become 100 degree celcius of water frm 25 degree celcius
Q=ms*del theta
Q=1*4200*(100-25)=?

Answer 3

Right?

Answer 4

don't we need to apply the Q=mHv equation?

Answer 5

or is it the same thing?

Answer 6

where did you get the 4200 from?

Answer 7

Now heat needed to become vapour
Q=mLv
Q=0.5*2.26*10^6=?

Answer 8

U hv to add those 2 Q

Answer 9

U will get total heat needed

Answer 10

thank u!

Answer 11

Anyway s is specific heat of watet

Answer 12

Welcome!!!"

Answer 13

I think u got it!!!!

Answer 14

Medal??????!!!!

Answer 15

after i calculate the Q, wont i have to cut my result in half because half is vaporized?

Answer 16

No!!!

Answer 17

I used m=0.5kg during vaporization

Answer 18

U know total mass 1 kg water become 100 degree celcius of water

Answer 19

yes

Answer 20

But 0.5 kg water become vapour frm 100 degree celcius

Answer 21

right! yay, i got it! thank you so much! u were extremely helpful!

Answer 22

Glad to help u!!!!