Question

Heat lost by cooling water = Q = -3470 Joules,
Mass of ice melted = 11.85 g

calculate the heat needed to melt 1 gram of ice (J/g)

Answer 1

Where are you stuck?

Answer 2

I just don't know how to incorporate those two values to figure out the answer?

Answer 3

is the question clear? or does it need more info? It's a lab experiment

Answer 4

Well, you know how much Energy is required to melt 11.85g of ice, 'cause that's the heat you lost (if I understood correctly). How would you find out how much heat is required to melt a certain fraction of that mass?

Answer 5

the Q was calculated using change in teperature (-8.8 degrees celsuis), and mass of water cooled (94.35 g) and 4.18 j/g C to be converted to joules

this whole things equals -3470 J

the mass of ice melted = 11.85 g

Could you explain a bit more how this is done? I'm not following.. thanks :)

Answer 6

are you saying that the amount of heat lost by cooling water = the amount of heat gained to melt 11.85 g of ice?

Answer 7

just devide total 3470 by 11.85
you will have per gram

Answer 8

that's it? no need to use Q = mc delta T or Q = mL ?

Answer 9

Yes, sort of.

I'm arguing, that there is something called the Conservation of Energy. It is a very important and fundamental law in physics and tells you, that Energy can neither be created nor destroyed. Meaning, if you are not exchanching energy with the "outside" - in our case, outside is everything not in your water-bowl - then, the total Energy in your system (the bowl with water and ice) does not change.

Furthermore, we'll assume the Energy that is exchanged via air to be neglectable.. that means: The energy we need to melt the ice has to come from somewhere: it is exactly the energy we 'loose' by cooling the water. The warmer water is giving it's heat to the ice to melt it.. and because heat can't just vanish, it hast to be the exact same amount..

Did you understand that?? :P

Answer 10

Yes.. I do

Answer 11

So I just divide 3470 by 11.85?

Answer 12

(Y)

Answer 13

the answer would be 292.82 J/g is this regarded as an accepted value of heat of fusion?

Answer 14

Yes, you will get the amount of heat needed _per_gram_, if you divide your total 'lost' energy by the number of grams you melted with it..

Answer 15

thanks.. well, there's one more thing. We need to calculate the percentage difference between an accepted value of heat of fusion (334 j/g) and the value we just calculated = 292.82 .. it would be -14.06% what does the negative sign indicate here? loss of heat?

Answer 16

This is acceptable, yeah.. the value according to literature is about 334 J/g ..

Answer 17

I'd say, you don't need that negative sign - the absolute value of the error is normally all you want to know. It indicates only, that your value is below the literature-value, not above it..

Answer 18

yeah it makes sense.. it's just that the equation given to calculate this is calculated value-accepted value / accepted value x 100

Answer 19

Well if it does not say "take the absolute value of that" or something, you can always write the negative sign. It's not wrong.. It's just not really of interest - usually ;)

Answer 20

Great.. thanks you so much @MuH4hA !! :) you've been of great help

Answer 21

You're very welcome :)