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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.

  • phys -

    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

  • phys -

    thank you...what I originally did was use the wrong specific heat for iron.

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