Question

A 50 kg man is running at a speed of 18 km $h^{-1}$. If all the kinetic energy of the man can be used to increase the temperature of water from ${20}^0$ C to ${30}^0$ C, how much water can be heated with this energy ?

Answer 1

K.E. of the man = $\frac{1}{2}m{V}^2$

= $\left(\frac{1}{2}\right)50\times 5^2$

= $25\times 25$ = 625 J

The amount of heat required to raise the temperature of water from ${20}^0$ C to ${30}^0$ C

Q= $ms\Delta \theta =m\times 4200\times$ (30-20)

Q= 42000 m

But, 42 $\times {10}^{3}m$ =K.E.$\Rightarrow 625$

m = $\frac{625}{42}\times {10}^{-3}$

= 14.88 $\times {10}^{-3}$ kg

$\cong$ 15 g