Thermal Properties of Matter

JEE Main level, single correct, medium

52 of 126

Question 52 of 126

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 m/s2g=10 \mathrm{~m} / \mathrm{s}^2, specific heat of water =4200 J/(kgK))\left.=4200 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\right)

Show answer and solution

Answer: (D) 0.48 K

mgh=msΔTΔT=ghs=10×2004200 K=1021 K\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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermal Properties of Matter (Calorimetry), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.