Okay, here's my full answer.
given specific heat of ice = .5 cal/g*K and specific heat water = 1 cal/g*K.
Latent heat of fusion for water is 334 J/g
1 calorie = 4.184 J
dQ_1 = Heat to bring ice to 0ºC
= 2.092 (J/g*K) * 8600 (g) * (0--6) (K) = 107,947 J
dQ_2 = Heat to melt ice.
= 334...
Maybe a typo on my part? Whatever I put into the homework site was accepted (i think the range is +/- 1% or so).
But can you explain why using the midpoint is only a rough approximation? Isn't it a linear graph when we don't consider the time involved?
Just thought I'd bump this quick.
Just finished a test involving topics ranging from the 2nd Law of Thermodynamics and Entropy to light/electromagnetic waves to refraction/reflection and mirrors/lenses.
With all those topics, fully 1/3 of the test involved an entropy problem similar to...
Found my error, found the right answer, but still wondering why it's clear why we can use the midpoint of dT.
Specific heat of ice \neq specific heat water. Whoops.
EDIT : Nevermind on that too. As long as we're not interested in the time it takes for the temperature to change, we...
First off, I really do appreciate the help!
However, a couple questions:
Is it really clear why we can use the midpoint? Sure, I get close to the right answer, but by using the midpoint we are assuming that T is changing at a constant rate, which isn't the case. While this implies the need...
After your first reply, that's what I was thinking.
However, how/when do I determine the T for each dQ?
ie, bringing the ice from -6 C to 0 C isn't a constant temperature, so how do I decide which value to use?
Well, the lake is transferring heat into the ice, and since the dQ1 + dQ2 = 0, the lake is contributing -3840.8 kJ.
If I understand this correctly, my above answer is actually the loss of entropy of the lake itself. I then need to find the gain in entropy of the ice. (that can't be the gain...
Homework Statement
A boater dumps 8.6 kg of ice from his cooler into the lake at the end of a fishing trip. The ice is at -6°C, and the lake is at 20.9°C. What is the change in entropy of the new lake-ice cube system after the ice cubes come to equilibrium with the lake?
Homework...
The equation you need here is E = \frac{kq}{r^2} * \hat{r}, where k is the permitivity of free space.
\hat{r} represents the radial unit vector from the point charge to the point where you are calculating the electric field, and shouldn't have a magnitude. (I think your * < x,y,z> is...
Another thing I found helpful when I took this class is, once you follow Doc Al's advice, to set up all the equations in a matrix. It helps make the algebra easier! :smile:
Thanks for the replies, I think I've got the right answer.
So the total power output of the sun is 4.47e26 W, and the SA of the new sphere is 4*pi*(1.5e11)^2 m^2 = 2.82743e23 m^2.
Dividing gives 1580.94 w*m^(-2). Sound right?
Homework Statement
Assume that the radiation emitted from the Sun moves radially outward from the Sun and that no radiation is absorbed between the Sun and Earth. How much energy in the form of radiation will fall per second on an area of 1 m2 on Earth, if that area is perpendicular to the...