Completely confused on this physics problem...?

Question

Ok, here's the problem...

"The specific heat capacity of ice is about 0.5cal/g x degrees celsius. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1-gram ice cube at absolute zero (-273 degrees celsius) to 1 gram of boiling water. How does this number of calories compare with the number of calories required to change the same gram of 100 degree celsius water to 100 degrees celsius steam?"

I've stared at this problem for a while, and I have no idea where to even start. Am i supposed to use the formula Q = mcDT? I have no idea! Please help!

Answer 1

You need these specific energy quantities:
Q(1): Energy needed to heat an ice cube from -273 C to 0 C
Q(2): Energy needed to turn 0 C ice to 0 C water
Q(3): Energy needed to heat 0 C water to 100 C water
Q(4): Energy needed to turn 100 C water to 100 C steam

Q(1) = mc*dT
Q(2) = m*H(f)
Q(3) = mc*dT
Q(4) = m*H(v)

Q(total) = Q(1) + Q(2) + Q(3) + Q(4)

Here's a definition of the variables in the "Q" equations:
m = 1 g (mass of H2O)
c = 0.5 cal/g*C
dT = change in temperature
H(f) = heat of fusion = 6.02 kJ/mol = 334 J (for 1 g of ice)
H(v) = heat of vaporization= = 40.7 kJ/mol = 2261 J (for 1 g of water)

You can convert from Joules to calories or calories to Joules using 1 calore = 4.184 Joules

The calories required to change the same gram of 100 degree celsius water to 100 degrees celsius steam is just Q(4) in the equation above.

Hope that helped!

Answer 2

Yes. The question's a bit ambiguous isn't it?

You can either read it as - use 1 cal for every 2 degrees change all the way up to 100C , ignoring the fact that you have to add the energy needed to melt the ice around 0C and that liquid water needs about 1cal for every 1 degree change from 0 to 100C then compare this figure with how much latent heat you have to add to vaporise it (Ashwin's method).

Or you can read it as , just assume this value up to 0C add in the latent heat needed to change it into liquid, use 1cal/degree C to 100C and then compare that sum with the amount of energy you need to add to vapourise it (Cirric's method and the only one that makes any sort of sense to me). In this sort of situation I'd trust that whoever set the question was more interested in the solution that was closer to physical reality, and in comparing two physically meaningful quantities to see if there were any general conclusions you might be able to draw than showing that you can do the math.

I'd also stick with standard pressure as it actually does mention boiling the water at 100C so there's no point in getting all fancy.
Best of Luck - Mike

Answer 3

Hi. Two problems. Get from -273 to melting point at 0. (.5 times 273). Then get from 0 to melting to steam at 100 deg C. Does this help?

Answer 4

Q=number of calories
m=mass of water
c=specific heat capacity
dT=change in temperature

so here,

Q=1x.5x373;
=186.5Cals

heat required to convert water to steam at 100 degrees is called latent heat of vaporisation, it has a different formula.

Answer 5

The phraseology of that query is greater perplexing than the concern itself. i'm uncertain which point of physics you're in, because of the fact there are some straight forward solutions and difficult solutions. it rather is actual asserting that the earth is spinning at a undeniable speed. by using default, something on its floor is travleing at that comparable speed besides. If, all of a audden, that spinning got here to a screeching halt, the products on the exterior could nevertheless be moving on the unique speed, tangent to the circle... Kinda like once you slam on your brakes, the persons in the motor vehicle nevertheless circulate forward. in any case, if the earth hit the brakes, people could fly off the exterior and hypothetically into area. remember, that the quicker the earth spins, the quicker its rotation and the greater speed products on the exterior have. The slower it spins, the slower its rotation and the less speed they have. you may desire to locate the element the place the speed of the products on the exterior exceeds the stress of gravity (i'm assuming it would go away the gravitational pull of the earth to proceed into area, otherwise products could return to earth, like a kicked soccer.