Sum

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

Concept: Anomalous Expansion of Water

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