A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal (g

Engineering

Physics

Calorimetry New

Question

A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal (g –ºC) (1 cal = 4.2 × 10⁷ ergs) close to :

119.2ºC

38.4ºC

83.3ºC

87.5ºC

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As per given condition

$\frac{1}{2} \times \frac{1}{2} {mv}^{2} = {(msΔT)}_{bullet}$

$Δt = \frac{V^{2}}{4 s}$

$= \frac{210 \times 210}{4 \times 4 .2 \times 0 .3 \times 1000} = 87 .5 º C$