Question

11. A 50 kg man is running at a speed of 18 km/h. 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 1

Answer:

10 ° may be but you do google

Answer 2

A 50 kg man is running at a speed of 18 km/h. If all the  kinetic energy of the man can be used to increase the temperature of water from 20°C to 30°C, 15 g of water can be heated with this energy.  

Explanation:

Step 1:

Given data  

m = 50 kg

v = 18 km/h = $18 \times \frac{5}{18}=5 \frac{m}{s}$

s = 4200 J/Kg-K

where

s  is Specific heat of the water,  

m is Mass of the man,  

v is Speed of the man,  

Step 2:

The Man's kinetic energy is provided by

$K=\frac{1}{2} m V^{2}$

$\begin{aligned}&\mathrm{K}=\left(\frac{1}{2}\right) 50 \times 5^{2}\\&K=25 \times 25=625 J\end{aligned}$

Let the mass of the heated water be M.

Step 3:

The amount of heat needed to raise the water temperature from 20 ° C to 30 ° C is provided by the following

                                $Q=m s \Delta T=M \times 4200 \times(30-20)$                                                                 Q = 42000 M

The question is as follows,

Q = K

42000 M = 625  

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

     $=14.88 \times 10^{-3}$

    $=15 g$