Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100
degrees Celsius.
Required:
Heat produced
Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules
per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat
produced in boiling 2 kg of water.
Answer:
is the drop in the water temperature.
Explanation:
Given:
- mass of ice,
- mass of water,
Assuming the initial temperature of the ice to be 0° C.
<u>Apply the conservation of energy:</u>
- Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.
<u>Now from the heat equation:</u>
......................(1)
where:
latent heat of fusion of ice
specific heat of water
change in temperature
Putting values in eq. (1):
is the drop in the water temperature.
Answer:
d)symmetry
Explanation:
Here is the complete question
In deriving the axial electric field for the ring-shaped charge distribution and the electric field from a long line of charge, the component perpendicular to the resulting field is zero because of what physical property? Answer a)integration b)superposition c)only net charge is important d)symmetry e) positive and negative charges cancel
Solution
The perpendicular components of the resulting electric field is zero because of symmetry since the charge is evenly distributed around the ring or line of charge. At one point, there is an electric field dE due to a small charge element dq and a corresponding electric field dE' due to a small charge element dq' opposite to dq.
Since the charges are symmetrical about the center, the horizontal (or perpendicular) components of the electric fields would cancel out causing the horizontal component of the resulting electric field to be zero.
So, the perpendicular component of the resulting electric field is zero due to symmetry.
Answer:
Explanation:
Given:
mass of the boat,
uniform speed of the boat,
rate of accumulation of water mass in the boat,
time of observation,
The mass of the boat after the observed time:
<u>Now using the conservation of momentum:</u>