Answer:
s = 589.3 m
Explanation:
Let the truck and car meet at a distance = s m
The truck is moving at constant velocity = v
so s= v * t ---------- (1)
car:
Vi = 0 m/s
a = 3.9 m/s²
s = Vi* t + 1/2 a t²
s= 0 * t + 1/2 a t²
s = 1/2 a t² ----------- (2)
compare equation (1) and equation (2)
s= v * t = 1/2 a t²
⇒ v * t = 1/2 a t²
⇒ t = 2 * v/ a
⇒ t = (2 * 33.9 )/ 3.9
⇒ t = 17. 38 s
Now
from equation (1)
s= v * t
s= 33.9 * 17.38
⇒ s = 589.3 m
When you heat a certain substance with a difference of temperature
the heat (energy) you must give to it is
where
is the specific heat of that substance (given in J/(g*Celsius))
In this case
Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a
difference of 1 degree Celsius = 1 degree Kelvin
Atomic mass = number of protons + number of neutrons = 4+5 = 9 amu
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.
