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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?
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Solution
Given:-
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
`K=1/2mV^2`
`K=(1/2)50xx5^2`
`K=25xx25=625 J`
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
`rArrM=625/42xx10^-3`
`=14.88xx10^-3`
=15 kg
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