<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
Answer:
because the partieals are cold are pushed away
Explanation:
Answer:
48.6 g
Explanation:
Given data:
Length of one side = 3 cm
Density of cube = 8.1 g/cm³
Mass of cube = ?
Solution:
First of all we will calculate the volume of cube.
V = length × height × width
V = 3 cm × 3cm× 3cm
V = 27cm³
Now we will determine the mass by using density formula.
d = m/v
d = density
m = mass
v = volume
8.1 g/cm³ = m/ 27 cm³
m = 8.1 g/cm³ ×27 cm³
m = 48.6 g
