A 25-kg iron cannonball is fired with speed 220 m/s. If it hits a concrete wall
ID: 1475543 • Letter: A
Question
A 25-kg iron cannonball is fired with speed 220 m/s. If it hits a concrete wall and half of its initial kinetic energy (KE = 1/2mv^2) is converted to internal energy in the cannonball (or heats the cannonball), then how much does the temperature of the cannonball increase The specific heat of iron is 450 J/kg-degreeC. 26.89 degreeC 22.24 degreeC 28.72 degreeC 33.25 degreeC 26.53 degree 23.49 degreeC Suppose 2.50 kg of 32 degreeC water is mixed with 9.30 kg of 65 degreeC water. If no heat is lost in this system, what is the final temperature of the water 46.36 degreeC 47.06 degreeC 58.01 degreeC 69.98 degreeC 54.30 degreeC 63.90 degreeC A hot (70.0 degreeC) metal object has a mass of 250-g and a specific heat of 250 J/kg degreeC. George drops the metal into a 60.0-g aluminum calorimeter containing 195-g of water at 14.0 degreeC. When equilibrium is reached, what will be the final temperature of the system c_aluminium = 900 J/kg degreeC, c_water = 4186 J/kg degreeC. 15.75 degreeC 17.74 degreeC 19.47 degreeC 17.75 degreeC 21.57 degreeC 21.72 degreeCExplanation / Answer
mass m = 25 kg
Speed v = 220 m/s
Initial kinetic energy K= (1/2) mv 2
= 0.5 x25 x220 2
= 605000 J
Internal energy E = (1/2) K
= 302500 J
This energy E is utilized to raise the temprature of the ball.
So, E = mC dt
From this dt = E / mC
Where C = Specific heat of iron = 450 J / kg o C
Substitute values you get dt = 26.89 o C
(13).Given m = 2.5 kg
M = 9.3 kg
t = 32 o C
t ' = 65 o C
Final temprature T = ?
Heat lost by mass M = Heat gain by mass m
MC(t ' - T ) = mC ( T - t )
M(t ' - T ) = m ( T - t )
9.3(65-T) = 2.5 (T - 32)
65 -T = 0.2688 (T-32)
= 0.2666 T - 8.602
1.2666 T = 73.602
T = 58.11 o C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.