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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.
\(\left(\right.\)Take \(g=10 \mathrm{~m} / \mathrm{s}^2\), specific heat of water \(\left.=4200 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\right)\)

  1. A 0.23 K
  2. B 0.36 K
  3. C 0.14 K
  4. D 0.48 K
Verified Solution

Answer & Solution

Correct Answer

(D) 0.48 K

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{mgh}=\mathrm{ms} \Delta \mathrm{T} \\ & \Delta \mathrm{T}=\frac{\mathrm{gh}}{\mathrm{s}}=\frac{10 \times 200}{4200} \mathrm{~K}=\frac{10}{21} \mathrm{~K}\end{aligned}\)
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