One end of a 2.35,m long and 2.0,cm radius al | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: Length $l = 2.35\,m$,Radius $r = 2.0\,cm = 0.02\,m$,Thermal conductivity $K = 235\,W\cdot m^{-1}K^{-1}$,Temperature difference $\Delta T =

Vedclass · Accepted Answer

$2.4\pi 	imes 10^{-6}$

Vedclass · Answer

(C) Given: Length $l = 2.35\,m$,Radius $r = 2.0\,cm = 0.02\,m$,Thermal conductivity $K = 235\,W\cdot m^{-1}K^{-1}$,Temperature difference $\Delta T = 20^{\circ}C$,Latent heat $L_f = \frac{10}{3} 	imes 10^5\,J\cdot kg^{-1}$.The rate of heat flow is given by $\frac{dH}{dt} = \frac{KA\Delta T}{l}$.The rate of ice melting is $\frac{dm}{dt} = \frac{1}{L_f} \frac{dH}{dt} = \frac{KA\Delta T}{l \cdot L_f}$.Area $A = \pi r^2 = \pi (0.02)^2 = 4\pi 	imes 10^{-4}\,m^2$.Substituting the values: $\frac{dm}{dt} = \frac{235 	imes 4\pi 	imes 10^{-4} 	imes 20}{2.35 	imes \frac{10}{3} 	imes 10^5}$.$\frac{dm}{dt} = \frac{235 	imes 4\pi 	imes 10^{-4} 	imes 20 	imes 3}{2.35 	imes 10^6} = \frac{100 	imes 4\pi 	imes 10^{-4} 	imes 60}{10^6} = \frac{2400\pi 	imes 10^{-4}}{10^6} = 2.4\pi 	imes 10^{-6}\,kg\cdot s^{-1}$.

Similar Questions

Two plates of the same area are placed in contact. Their thicknesses as well as their thermal conductivities are in the ratio $2:3$. The outer surface of one plate is maintained at $10^{\circ} C$ and that of the other at $0^{\circ} C$. The temperature at the common surface is (in $^{\circ} C$)

The end $A$ of a rod $AB$ of length $1\,m$ is maintained at $80\,^{\circ}C$ and the end $B$ at $0\,^{\circ}C$. The temperature at a distance of $60\,cm$ from the end $A$ is......... $^{\circ}C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module