A coil of wire of resistance 50,Omega is embe | vedclass

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

Question

(D) The heat produced by the coil per second is given by $H = \frac{V^2}{R}$.Since the heat is used to melt the ice,we have $H = \frac{m}{t} 	imes L$

Vedclass · Accepted Answer

$2.625\,gm$

Vedclass · Answer

(D) The heat produced by the coil per second is given by $H = \frac{V^2}{R}$.Since the heat is used to melt the ice,we have $H = \frac{m}{t} 	imes L$,where $L$ is the latent heat of fusion of ice $(L = 80\,cal/g = 336\,J/g)$.To convert electrical energy in Joules to calories,we use the conversion factor $J = 4.2\,J/cal$.The rate of melting is $\frac{m}{t} = \frac{V^2}{R 	imes L 	imes 4.2}$.Substituting the values: $\frac{m}{t} = \frac{210^2}{50 	imes 80 	imes 4.2} = \frac{44100}{16800} = 2.625\,g/s$.

$A$ coil of wire of resistance $50\,\Omega$ is embedded in a block of ice. If a potential difference of $210\,V$ is applied across the coil,the amount of ice melted per second will be

Similar Questions

$A$ certain charge liberates $0.8\, gm$ of ${O_2}$. The same charge will liberate how many $gm$ of silver?

The cold junction of a thermocouple is at $10\,^{\circ}\text{C}$. When the temperature of the hot junction is $530\,^{\circ}\text{C}$,no thermo-emf is produced. What is the neutral temperature in $^{\circ}\text{C}$?

Total $emf$ produced in a thermocouple does not depend on

$A$ battery is charged by a $15\,V$ battery for $8\,h$ at a current of $10\,A$. When this battery discharges, it provides a current of $5\,A$ for $15\,h$. The average terminal voltage during discharge is $14\,V$. What is the $watt-hour$ efficiency of this battery in $\%$?

As the temperature of the hot junction of a thermocouple is increased (while the cold junction is at a constant temperature), the thermo $e.m.f.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module