Question

A 50$kg$ man is running at a speed of 18$km{h}^{-1}$ .If all the kinetic energy of the man can beused to increase the temperature of waterfrom ${20}^{\circ }C$ to ${30}^{\circ }C,$ how much water can beheated with this energy?

• A. 15$g$
• B. 20$g$
• C. 30$g$
• D. 40$g$

Answer 1

The correct option is A 15$g$

Given$:$

$v=18km/h=5m/s$

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

So kinetic energy$:$

$K=\frac{1}{2}\times 50\times {\left(5\right)}^2=625J$

Now,

$\Delta T={30}^{0}C-{20}^{0}C={10}^{0}C$

$\left(\Delta Tin{}^{0}Cisthesameas\Delta TinK\right)$

So$,$ we use$:$

$Q=m_{H_{2}O}{C}_{H_{2}O}\Delta T$

$625=m_{H_{2}O}\times 4.18\times 10$

$m_{H_{2}O}=14.9g\approx 15g$

Hence,

option $\left(A\right)$ is correct answer.