Complete question:
Two cars collide head on while each is traveling at 90 km/h. Suppose all their kinetic energy is transformed into the thermal energy of the wrecks. What is the temperature increase of each car?
you may assume that the specific heat capacity of each car is that of
iron..
Answer:
The temperature increase of each car is 0.68 k
Explanation:
Given;
velocity of each car, v = 90 km/h
velocity of each car (in m/s) = (90 x 10)/36 = 25 m/s
Kinetic energy of each car, KE = ¹/₂ mv²
Total Kinetic energy of the two cars = 2 x ¹/₂ mv² = mv²
Thermal energy = mcΔT
Total thermal energy of the the two cars = 2 x mcΔT = 2mcΔT
Suppose all their kinetic energy is transformed into the thermal energy of the wrecks, then; 2mcΔT = mv²
divide both sides by m,
2cΔT = v²
where;
c is the specific heat of iron = 460.548 J/kg.K
ΔT is the temperature increase of each car
v is the velocity of each car = 25 m/s
Substitute these values and calculate ΔT
ΔT = v²/2c
ΔT = (25)²/(2 x 460.548)
ΔT = 0.679 k = 0.68 k
Therefore, the temperature increase of each car is 0.68 k
The temperature increase of each car is 0.68 K.
Given data:
The speed of each car is, [tex]v= 90 \;\rm km/h =25 \;\rm m/s[/tex].
The kinetic energy of each car is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
Here, m is the mass of each car. Then total kinetic energy is,
[tex]KE_{T}=2KE\\\\KE_{T}=2 \times \dfrac{1}{2}mv^{2}\\\\KE_{T}=mv^{2}[/tex]
And the thermal energy of each car is,
Q = mcΔT
Total thermal energy of the the two cars
Q' = 2 x mcΔT
Q = 2mcΔT
Suppose all their kinetic energy is transformed into the thermal energy of the wrecks, then;
[tex]KE_{T} = Q'[/tex]
mv² = 2mcΔT
2cΔT = v²
here,
c is the specific heat of iron and its value is 460.548 J/kg.K
ΔT is the temperature increase of each car
Substitute these values and calculate ΔT
ΔT = v²/2c
ΔT = (25)²/(2 x 460.548)
ΔT = 0.679 K
ΔT = 0.68 K
Thus, we can conclude that the temperature increase of each car is 0.68 K.
Learn more about the kinetic energy here:
https://brainly.com/question/12669551
