The Astonishing Heat Capacity of Snow: Explained

AI Thread Summary
The discussion centers on understanding why 30,000 kg of snow at 0°C contains more heat energy than 1 mL of liquid water at 100°C. It emphasizes that temperature measures average thermal energy per molecule, not total thermal energy. The concept of heat capacity is crucial, as larger quantities of a substance at the same temperature can store more energy. To compare the two, one can use the heat of fusion and the formula dq = mcΔT to calculate the energy required to change the states of both the snow and the water. This analysis reveals that the sheer mass of snow significantly contributes to its total heat energy despite its lower temperature.
bandrews789
Messages
2
Reaction score
0
Explain how it is possible for a 30,000 kg of snow at 0 deg C to contain more heat energy than 1 mL of liquid water at 100 deg C. (Assume a pressure of one atmosphere.) I am unsure of where to start this? Help Is there a formula?
 
Last edited:
Physics news on Phys.org
This is more of a conceptual thing: temperature is not thermal energy. Anything that is above absolute zero (zero Kelvins = -273 celcius) has thermal energy. Temperature is a measure of the average thermal energy per molecule in an object.

THink this way: what has more energy, a small cup of water at 100 C or a large barrel of water at 100 C? To put it another way, which one, the cup or barrel, needs more energy to get the water from 50 C to 100 C?
 
Explain how it is possible for a 30,000 kg of snow at 0 deg C to contain more heat energy than 1 mL of liquid water at 100 deg C. (Assume a pressure of one atmosphere.) I am unsure of where to start this? Help Is there a formula?

apply a phase diagram (and assume that heat capacity of the constituents of snow is somewhat identical to that of H20), apply the heat of fusion value to completely "freeze" the snow to find the heat energy of the process...bring the 1mL of water at 100C to the same state; find the corresponding heat energy (100C, apply dq=mcdT, bring it to 0C and apply the heat of fusion value).
 
Thank you for your assistance in this matter
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top