A copper bullet is traveling at 260 m/s and impacts a solid block(ideal thermal

Question

A copper bullet is traveling at 260 m/s and impacts a solid block(ideal thermal insulator) and comes to rest at its center. The temperature of the bullet, jut prior to impact is25ºC. The mass of the bullet is 0.013 kg and thespecific heat capacity is 385 J/kg*K. Assume the impact iscompletely inelastic. a)What is the final temperature of the bullet just afterimpact? b)The hot copper is then quickly placed into a thermallyisolated container holding 0.010 kg of water. What is thefinal temperature of the water when it reaches thermal equilibriumwith the bullet? Assume that no heat is lost during thetransfer of the bullet and the water does not evaporate. c)How fast would the bullet have needed to be traveling atimpact in order to completely melt assuming the melting point ofcopper is 1358ºC and the latent heat of fusion is 2.09x105J/kg? a)What is the final temperature of the bullet just afterimpact? b)The hot copper is then quickly placed into a thermallyisolated container holding 0.010 kg of water. What is thefinal temperature of the water when it reaches thermal equilibriumwith the bullet? Assume that no heat is lost during thetransfer of the bullet and the water does not evaporate. c)How fast would the bullet have needed to be traveling atimpact in order to completely melt assuming the melting point ofcopper is 1358ºC and the latent heat of fusion is 2.09x105J/kg?

Explanation / Answer

(a) KE of bullet = (1/2) mv2 = (1/2) * 0.013 * 2602 =   439.4 Joules . All of the KE is transformed to heat, so... .            439.4 = m c T . T = 439.4 / 0.013 * 385 =   87.79 K     the temp increases bythis much so... .    final temp = init temp +T = 25 + 87.8 =   112.8 degC . (b)   Now...    heat lost bybullet = heat gained by water .            m c T for bullet = m c T forwater .           0.013 *385 * (112.8 - Tf ) = 0.01 * 4186 * (Tf - Ti ) . NOTICE:   you did not provide the initialtemp of the water, so part b cannot be answered. If you have theinitial temp of the water, plug it into the above equation andsolve for Tf. Please carefully check the infogiven in your problem. . (c)    to melt the bullet... .     KE = heat to melt bullet .    (1/2) m v2 =   m cT   + m L .    v2 = 2 cT   + 2L = 2 * 385 *(1358-25) + 2*209000 = 1444410 . v =   1202 m/s

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