Class 11Physics10-1.Thermometry, Thermal Expansion and CalorimetryMix Examples-Thermometry, Thermal Expansion and Calorimetry
MediumThe top of a lake gets frozen at a place where the surrounding air is at a temperature of $-20\,^oC$. Then
- AThe temperature of the layer of water in contact with the lower surface of the ice block will be at $0\,^oC$ and that at the bottom of the lake will be $4\,^oC$
- BThe temperature of water below the lower surface of ice will be $4\,^oC$ right up to the bottom of the lake
- CThe temperature of the water below the lower surface of ice will be $0\,^oC$ right up to the bottom of the lake
- DThe temperature of the layer of water immediately in contact with the lower surface of ice will be about $-20\,^oC$ and that of water at the bottom will be $0\,^oC$
Explore More
Similar Questions
$A$ block of steel of mass $2 \,kg$ slides down a rough inclined plane of inclination $\sin ^{-1}\left(\frac{3}{5}\right)$ at a constant speed. Assuming that the mechanical energy lost due to friction is used to increase the temperature of the block, calculate the rise in temperature of the block as it slides through $80 \,cm$. (Specific heat capacity of steel $= 420 \,J kg^{-1} K^{-1}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $^{\circ} C$)
The height of Victoria Falls is $63 \ m$. What is the difference in temperature of water at the top and at the bottom of the fall? (In $^{\circ}C$)
[Given $1 \ cal = 4.2 \ J$ and specific heat of water $= 1 \ cal \ g^{-1} \ ^{\circ}C^{-1}$]
[Given $1 \ cal = 4.2 \ J$ and specific heat of water $= 1 \ cal \ g^{-1} \ ^{\circ}C^{-1}$]
Calculate the amount of heat (in calories) required to convert $5\,g$ of ice at $0^{\circ}C$ to steam at $100^{\circ}C$.
Medium
View SolutionHeat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 \,g/s$. It melts completely in $100 \,s$. The rate of rise of temperature of the resulting water thereafter will be ............ $^{\circ}C/s$.
Medium
View SolutionThe heat extracted out of $x \ \text{g}$ of water initially at $50^\circ \text{C}$ to cool it down to $0^\circ \text{C}$ is sufficient to evaporate $(1000 - x) \ \text{g}$ of water also initially at $50^\circ \text{C}$. The value of $x$ (closest integer) is . . . . . . . (Take latent heat of water $L = 2256 \ \text{kJ/kg}$, specific heat capacity of water $c = 4200 \ \text{J/kg}\cdot \text{K}$)
DifficultJEE MAIN 2026
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo