Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
<span>D. Pressure increases with increasing depth.
This occurs because there is more weight above you to increase the pressure.
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Answer:
Mass of the oil drop, 
Explanation:
Potential difference between the plates, V = 400 V
Separation between plates, d = 1.3 cm = 0.013 m
If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :
, e is the charge on electron
E is the electric field, 
So, the mass of the oil drop is 
. Hence, this is the required solution.
Answer:
is there an equasion it gives you?
Explanation:
need a little more info.
Letter na that would be the answer
