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°C to 30°C, how much water can be heated with this energy?

Answer 1

Given:
Mass of the man, m = 50 kg
Speed of the man, v = 18 km/h = $18\times \frac{5}{18}=5\mathrm{m}/\mathrm{s}$

Kinetic energy of the man is given by
K$=\frac{1}{2}m{\mathrm{V}}^2$

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

Specific heat of the water, s = 4200 J/Kg-K
Let the mass of the water heated be M.

The amount of heat required to raise the temperature of water from 20°C to 30°C is given by
Q = msΔT = M × 4200 × (30 − 20)
Q = 42000 M

According to the question,
Q = K

42000 M = 625

$$\begin{gathered} \Rightarrow M=\frac{625}{42}\times {10}^{-3} \\ =14.88\times {10}^{-3} \\ =15\mathrm{g} \end{gathered}$$