The velocity of a 42 kg celestial body reduces from `20 km cdot "min"^(-1)` to `5 km cdot "min"^(-1)` due to its passage through the earth's atmosphere. Find out the heat produced in calorie. `(J = 4.2 xx 10^(7) erg cdot cal^(-1))`

Explanation

Initial velocity of the celestial body,
`u = 20 km cdot "min"^(-1) = (20 xx 1000)/(60) m cdot s^(-1)`,
Final velocity of the celestial body,
`v = 5 km cdot "min"^(-1) = (5 xx 1000)/(60) m cdot s^(-1)`
Work done, W = Change in kinetic energy
`=1/2 mu^(2) - 1/3 mv^(2) = 1/2 m(u^(2) - v^(2))`
`=1/2 xx 42 xx ((1000)/(60))^(2) (20^(2) - 5^(2))`
`= (42 xx 10^(4) xx 375)/(2 xx 36) "joule" `
`:.` Heat produced
`H = W/J = (42 xx 10^(4) xx 375)/(2 xx 36 xx 4.2)`
`[J = 4.2 xx 10^(7) erg cdot cal^(-1) = 4.2 J cdot cal^(-1)]`
`=5.2 xx 10^(5) cal`.