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Karnataka Board PUCPUC Science Class 11

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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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?

Sum
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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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Anomalous Expansion of Water
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Chapter 25: Calorimetry - Exercises [Page 47]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 25 Calorimetry
Exercises | Q 11 | Page 47
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