5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please

Question

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the

step by step please

Susmita · Answer

There is a formula for freezing of a lake.I am writing it down here.If you need to know how that formula came ask me here. Time taken=(dL/2KT)×(x 2 2 - x 1 2 ) where d=density L=latent heat K=thermal conductivity T= temperature in degree centigrade x 1 =initial thicknes of ice sheet formed x 2 = final thickness of ice sheet formed. In the given problem, T=|263-273|=10 x 1 =0 and x 2 =0.5 meter and you know d,L,K.

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Guest

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of 263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and density ρ = 0.9 × 103 kgm-3. Take the step by step please

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Thermal Physics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Guest

IIT JEE Courses

One Year IIT Programme

Two Year IIT Programme

Crash Course

NEET Courses

One Year NEET Programme

Two Year NEET Programme

One to One

School Board

Live Online classes

Class 11 & 12

Class 9 & 10

Class 6, 7 & 8

Test Series

JEE test series

NEET test series

C.B.S.E test series

Complete Self Study Packages

FULL COURSE

For class 12th

For class 11th

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of 263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and density ρ = 0.9 × 103 kgm-3. Take the step by step please

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 anddensity ρ = 0.9 × 103 kgm-3. Take thestep by step please

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Thermal Physics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please