Water falls from a height of 200~m$200\mathrm{~m}$ into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take g = 10~m/s²$g = 10\mathrm{~m/s}^{2}$, specific heat of water = 4200~J/(kg· K)$= 4200\mathrm{~J/(kg\cdot K)}$)
A.0.23~K$0.23\mathrm{~K}$
B.0.36~K$0.36\mathrm{~K}$
C.0.14~K$0.14\mathrm{~K}$
D.0.48~K$0.48\mathrm{~K}$
Solution & Explanation
Related Formula
Δ U = mgh and Q = msΔ T$$\Delta U = mgh \quad \text{and} \quad Q = ms\Delta T$$
By conservation of energy (assuming all potential energy goes into heating the water):
mgh = msΔ T Δ T = (gh)/(s)$$mgh = ms\Delta T \implies \Delta T = \frac{gh}{s}$$
where,
g$g$ = acceleration due to gravity
h$h$ = height of the fall
s$s$ = specific heat of water
Δ T$\Delta T$ = rise in temperature
Core Logic
Given parameters:
h = 200~m$h = 200\mathrm{~m}$
g = 10~m/s²$g = 10\mathrm{~m/s}^{2}$
s = 4200~J/(kg· K)$s = 4200\mathrm{~J/(kg\cdot K)}$
Substitute the values to find Δ T$\Delta T$:
Δ T = (10 × 200)/(4200) = (2000)/(4200)$$\Delta T = \frac{10 \times 200}{4200} = \frac{2000}{4200}$$Δ T = (10)/(21) ≈ 0.476~K ≈ 0.48~K$$\Delta T = \frac{10}{21} \approx 0.476\mathrm{~K} \approx 0.48\mathrm{~K}$$
Pattern Recognition
Sees: "Water falling from height h$h$ raises temperature" → Mass cancels out. Δ T = (gh)/(s)$\Delta T = \frac{gh}{s}$.
Shortcut: Always use SI units (swater = 4200~J/kg· K$s_{\text{water}} = 4200\mathrm{~J/kg\cdot K}$ is given; if given in cal/g·^° C$\mathrm{cal/g\cdot^\circ C}$, convert using 1~cal = 4.184~J$1\mathrm{~cal} = 4.184\mathrm{~J}$). ✓
Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Work, Energy and Power
Keywords:#Water fall temperature rise calculation#JEE Main 2025 Evening Q5#Thermodynamics specific heat of water#Mechanical equivalent of heat conservation
More Thermodynamics Previous-Year Questions — Page 8
Qjee_main_2024_31_jan_morningIsobaric Process
The given figure represents two isobaric processes for the same mass of an ideal gas, then
A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.
A.P₂≥ P₁$\mathrm{P}_2\geq \mathrm{P}_1$
B.P₂ > P₁$\mathrm{P}_2 > \mathrm{P}_1$
C.P₁ = P₂$\mathrm{P_1 = P_2}$
D.P₁ > P₂$\mathrm{P}_1 > \mathrm{P}_2$
Solution
Related Formula
PV = nRT$PV = nRT$
Core Logic
From the Ideal Gas Law:
V = ((nR)/(P)) T$$V = \left(\frac{nR}{P}\right) T$$
In a V-T$V-T$ graph, the equation of the line represents y = mx$y = mx$, where the slope m$m$ is:
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
