A lead bullet of 10 g travelling at 300 m s- 1 strikes against a block of wood comes to rest. Assuming 50 % of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = 150 J kg- 1 k- 1 )

Question

A lead bullet of 10 g travelling at 300 m s- 1 strikes against a block of wood comes to rest. Assuming 50 % of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = 150 J kg- 1 k- 1 ) (A) 100℃ (B) 125℃ (C) 150℃ (D) 200℃

Tardigrade Admin · Accepted Answer

Here, m=10 g = 10- 2 kg v=300 m s- 1, Δ θ =?, C=150 text J text k textg- 1 textK- 1 Q=(50/100)((1/2) m v2)=(1/4)× (10)- 2(300)2=225 J From Q = m C Δ θ θ =(Q/C m)=(225/150 × 10- 2)=150 â

Q. A lead bullet of $10 g$ travelling at $300 m s^{- 1}$ strikes against a block of wood comes to rest. Assuming $50%$ of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = $150 J k g^{- 1} k^{- 1}$ )

$100℃$

$125℃$

$150℃$

$200℃$

Here, $m = 10$ g = $1 0^{- 2}$ kg
$v = 300 m s^{- 1}, Δ θ = ?, C = 150$ $J k g^{- 1} K^{- 1}$
$Q = \frac{50}{100} (\frac{1}{2} m v^{2}) = \frac{1}{4} \times (10)^{- 2} (300)^{2} = 225$ J
From $Q = m C Δ θ$
$θ = \frac{Q}{C m} = \frac{225}{150 \times 1 0 ^{- 2}} = 150 ℃$

Q. A lead bullet of 10g travelling at 300ms−1 strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = 150Jkg−1k−1 )

100℃

125℃

150℃

200℃