A 50 kg man is running at a speed of 18 km h^ -1 . If the entirety...

Velocity of the man, $v = 18 \text{ km h}^{-1} = 18 \times \dfrac{5}{18} \text{ m s}^{-1} = 5 \text{ m s}^{-1}$

Kinetic energy of the man, $K = \dfrac{1}{2}mv^2$

$K = \dfrac{1}{2} \times 50 \times (5)^2 = 625 \text{ J}$

This kinetic energy is utilized to heat the water. The heat energy required is given by $Q = m_w s \Delta T$

Given \Delta T = 30^{}^{{}^{\circ}}\text{C} - 20^{}^{{}^{\circ}}\text{C} = 10^{}^{{}^{\circ}}\text{C} and specific heat of water s \approx 4.18 \text{ J g}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

Equating kinetic energy to heat energy:

$625 = m_w \times 4.18 \times 10$

$m_w = \dfrac{625}{41.8} \text{ g} \approx 14.95 \text{ g}$

Rounding to the nearest integer, the mass of water that can be heated is $15 \text{ g}$.