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 fragmento must have entered the atmosphere is

Options

A. `0.45km//s`
B. `1.32km//s`
C. `2.32km//s`
D. zero

Answer

B

Explanation

The entire kinetic energy of the fragment is changed to heat. Expressing mass m in kilograms everywhere, we have
`(1)/(2)mv^2=[m(30)+m(0.11)(1535^@+100^@)]4184`
`v^2=(8368)[30+180]=1.76xx10^6`
Hence, `v=(sqrt(1.76))xx10^3=1.32 km//s`