m1 = 10.9•10^-3 kg, v =400 m/s, t(o) = 20oC, m2 = 0.4 kg,.
specific heat of the iron and wood, respectively,
c1 = 444 J/kg, and c2 = 1700 J/kg.
(a) m1•v^2/2 = 10.9•10^-3•(400)^2/ 2 =872 J = QQ/
Q = m1•c1•Δt,
Δt = Q/ m1•c1 = 872/10.8•10^-3•444 = 180oC.
t(new) = 20oC + 180oC =200oC.
(b)
m1•c1•(200-t) = m2•c2•(t-20),
t = (m1•c1•200 + m2•c2•20)/( m1•c1+ m2•c2) =
(10.9•10^-3•444•200 + 0.4•1700•20)/(10.9•10^-3•444+0.4•1700) =
= 21.27oC
A 10.9-g iron bullet with a speed of 4.00 102 m/s and a temperature of 20.0° C is stopped in a 0.400-kg block of wood, also at 20.0° C.
(a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature T. Calculate T, assuming no heat is lost to the environment.
2 answers
thank you...what I originally did was use the wrong specific heat for iron.
1 answer