NEET-XII-Physics
25: Calorimetry
- #11A 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?Ans : 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}``
`` K=\left(\frac{1}{2}\right)50\times {5}^{2}``
`` K=25\times 25=625\,\mathrm{\,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
`` \Rightarrow M=\frac{625}{42}\times {10}^{-3}``
`` =14.88\times {10}^{-3}``
`` =15\,\mathrm{\,g\,}``
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