A lead bullet reaches a maximum speed of 2500 ft/s before it hits a wall. The he

Question

A lead bullet reaches a maximum speed of 2500 ft/s before it hits a wall. The heat capacity of lead is 0.13 kJ/kg.K and the heat of melting is 22.4 kJ/kg. Lead melts at 328 C. You can assume that the temperature of the bullet before it hits the wall is 25 °C even though it is likely to be higher due to air friction. What is the kinetic energy (J/g) of the bullet? What is the energy (J/g) required to heat the bullet to its melting point and then melt it? Will the bullet melt if half of the kinetic energy is converted to internal energy? a. b. c.

Explanation / Answer

The question is to find Kinetic energy per gram of of lead, Kinetic energy is given as ,

KE - 1/2 * mv2

The units of KE are in joule, m in Kg and v in m/sec To get the values in terms of per mass units, we have to divide the equation by the mass of the substance of which KE is being calculated. On dividing by m,we get

KE/m = 1/2 * v2

Thus units left side have changed to J/Kg. In order to change to J/g , use conversion 1 Kg = 1000 g, in the above equation.

KE/ m(in grams) = KE / 1000 * m (in Kgs) = 1/2000 * v2

Alternatively, using basis system. Let basis be 1 gram of lead. i,e the bullet weight.. So we have,

m = 1 gram = 0.001 Kg

KE = 1/2 * 0.001 * v2 = 1/2000 * v2

This is KE f or 1 gram of lead. So KE per gram of lead =  1/2000 * v2.

For 7.5 gram as basis , KE = 1/2 * 7.5/1000 * v2 = 7.5/2000 * v2

Now for 7.5 g, KE =  7.5/2000 * v2 , Then KE per gram of lead = [7.5/2000 * v2]/7.5 = 1/2000 * v2 ( Direct variation)

a) KE per gram = 1/2000 * v2  

v = 2500 ft/sec = 2500 * 0.3048 m/sec = 762 m/sec

KE per gram = 7622 / 2000 = 29.0322 J/g

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