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?
Mass of the man, m = 50 kg
Speed of the man, v = 18 km/h = `18xx5/18=5"m/s"`
Kinetic energy of the man is given by
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