12.5 times 14 and convert to meters its 1.75 meters per second
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
Answer:
3.78075x10^6 kg x meter^2/ second^2
Explanation:
Kinetic energy K= 1/2 x mass x velocity^2
Answer:
- 0.8333 m/s^2
Explanation:
the equation for calculating acceleration that I have used is
(V2-V1)/t
where V2 is the final recorded velocity and V1 is the initial velocity and t is the duration of the acceleration.
which you plug the numbers in (15 - 25)/12 = 10/12 and because you are slowing down, it should be negative
I am in highschool honors physics so I am not college leveled, I might be wrong but I hope this is useful,