Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
Answer:
is time required to heat to boiling point form initial temperature.
Explanation:
Given:
initial temperature of water, 
time taken to vapourize half a liter of water, 
desity of water, 
So, the givne mass of water, 
enthalpy of vaporization of water, 
specific heat of water, 
Amount of heat required to raise the temperature of given water mass to 100°C:
Now the amount of heat required to vaporize 0.5 kg of water:
where:
mass of water vaporized due to boiling
Now the power rating of the boiler:
Now the time required to heat to boiling point form initial temperature:
I believe the correct answer from the choices listed above is the second option. The scientific notation of the measurement 0.00000000062 kg would be <span>6.2 x 10^-10 kg. Scientific notation is used to express too large and too small values of numbers. Hope this helps. Have a nice day.</span>
Answer:
Explanation:
Both these questions are based on the Universal Law of Gravitation, which is given by :
F = Gm1m2 / r²
2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5
F = 6.67 x 10⁻⁵ x 8 / 1.5
F = 35.57 x 10⁻⁵ N
3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴
F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29
F = 63.13 x 10⁻³ N
Answer:
Choice A
Explanation:
The lower the point the higher the kinetic energy because Mechanical energy is conserved and the Gravitational Potential Energy gets lower when the height is lower
