Question

a bullet is fired with 125m/s get embedded in a target of mass 2.74 kg as a result 80% of its energy when convert into heat energy temperature rises by 50 degree Celsius find specific heat of bullet ​

Answer 1

☆Concept:

• As the bullet is fired with speed it will have some kinetic energy.

• Due to resistive forces in atmosphere, some energy will be lost and get converted into heat.

☆Formula:

• $\bf{ KE = \dfrac{1}{2}mv²}$

• $\bf{ Heat= ms \triangle{T}}$

('s' is specific heat; 'T' is temperature)

☆Solution:

• KE = 1/2mv² = 1/2×2.74×(125)²

• KE = 1.37×15625

• KE = 21406.25 J

According to ques:

》KE × 80% = heat

》21406.25×80/100 = $\ ms \triangle{T}$

》17125 = 2.74 × s × 50

》$\ s = \dfrac{17125}{2.74×50}$

》$\ s = \dfrac{17125}{137}$

》s = 125 J/kg-kelvin.

☆ so the specific heat of bullet is 125.