An iron rocket fragment initially at -100° C enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is 0.11 kcal / kg ° C its melting point is 1535° C and the latent heat of fusion is 3 kcals / kg. The minimum velocity with which the fragment must have entered the atmosphere is

Question

An iron rocket fragment initially at -100° C enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is 0.11 kcal / kg ° C its melting point is 1535° C and the latent heat of fusion is 3 kcals / kg. The minimum velocity with which the fragment must have entered the atmosphere is (A) 0.45 km / s (B) 1.32 km / s (C) 2.32 km / s (D) zero

Tardigrade Admin · Accepted Answer

The entire kinetic energy of the fragment is changed to heat. Expressing mass

m

in kilograms everywhere, we have

\frac{1}{2} m v^{2} = [m (30) + m (0.11) (153 5^{\circ} + 10 0^{\circ})] 4184

v^{2} = (8368) [30 + 180] ≃ 1.76 \times 1 0^{6}

\Rightarrow v = (1.76) \times 1 0^{3} = 1.32 km / s

$0.45 km / s$

$1.32 km / s$

$2.32 km / s$

zero

The entire kinetic energy of the fragment is changed to heat. Expressing mass $m$ in kilograms everywhere, we have
$\frac{1}{2} m v^{2} = [m (30) + m (0.11) (153 5^{\circ} + 10 0^{\circ})] 4184$
$v^{2} = (8368) [30 + 180] ≃ 1.76 \times 1 0^{6}$
$\Rightarrow v = (1.76) \times 1 0^{3} = 1.32 km / s$

0.45km/s

1.32km/s

2.32km/s

zero

The entire kinetic energy of the fragment is changed to heat. Expressing mass m in kilograms everywhere, we have 21​mv2=[m(30)+m(0.11)(1535∘+100∘)]4184 v2=(8368)[30+180]≃1.76×106 ⇒v=(1.76​)×103=1.32km/s

$0.45 km / s$

$1.32 km / s$

$2.32 km / s$