Question

A 30g bullet at v=900m/s strikes a 1kg soft iron target stopping inside the iron [c=490J/kg°C. How much will the temperature of the iron increase? Ignore the heat that will be shared with the bullet

A) 25°C
B) 24795°C
C) 826°C
D) 82653°C
show your full work please

Answer 1

ΔT = 25°C

Step-by-step explanation:

Given that.

The mass of a bullet, m₁ = 30 g = 0.03 kg

The speed of the bullet, v = 900 m/s

Mass of soft iron, m₂ = 1 k

The specific heat of iron, c=490J/kg°C

We need to find the increase in temperature of iron. using the conservation of energy,

Kinetic energy = heat absorbed

`(1)/(2)m_1v^2=m_2c\Delta T\\\\\Delta T=((1)/(2)m_1v^2)/(m_2c)\\\\\Delta T=((1)/(2)* 0.03* 900^2)/(1* 490)\\\\=24.79^(\circ) C\\\\or\\\\\Delta T=25^(\circ) C`

So, the correct option is (A).