A `50 kg` man is running at a speed of `18 kmh^(-1)` If all the kinetic energy of the man can be used to increase the temperature of water from `20^(@)C` to `30^(@)C`.How much water can be heated with this energy?

Answer

A

Explanation

K.E. of the mass `== (1)/(2) mV^(2)`
`= ((1)/(2)) 50 xx 5^(2)`
`= 25 xx 25 625J`
The amount of heat required to raise the temperature of water from `20^(@)C to 30^(@)C `
`msDelta theta= m xx(30 - 20)`
`4200m`
`But, `42 xx 10^(3) m = 625`
`rArr m = (625)/(42) xx 10^(-3)`
` = 14.88 xx 10^(-3) kg`
`15 g`