An electric kettle takes $4\, A$ current at $220\, V$. How much time will it take to boil $1\, kg$ of water from room temperature $20\,^\circ C$ ? The temperature of boiling water is $100\,^\circ C$ .................. $minutes$
Medium
Download our app for free and get started
(b) Electric power consumed by kettle $P = 220 \times 4\,W$
Heat required
$H = 1000 \times 1\,(100 - 20) = 1000 \times 80\,\,cal$$ = 4200 \times 80\,\,J$
$P = \frac{H}{t} \Rightarrow \,H = P \times t$
$220 \times 4 \times t = 4200 \times 80$ $ \Rightarrow $ $t = 6.3\,\,$ $minutes$
Heat required
$H = 1000 \times 1\,(100 - 20) = 1000 \times 80\,\,cal$$ = 4200 \times 80\,\,J$
$P = \frac{H}{t} \Rightarrow \,H = P \times t$
$220 \times 4 \times t = 4200 \times 80$ $ \Rightarrow $ $t = 6.3\,\,$ $minutes$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionTwo wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be
- 2For a cell terminal $P.D.$ is $2.2\;V$ when circuit is open and reduces to $1.8\;V$ when cell is connected to a resistance of $R = 5\,\Omega $. Determine internal resistance of cell $(r)$ is then ........ $\Omega$View Solution
- 3The circuit diagram shown consists of a large number of element (each element has two resistors $R_1$ and $R_2$). The resistance of the resistors in each subsequent element differs by $a$ factor of $K = 1/2$ from the resistance of the resistors in the previous elements. The equivalent reistance between $A$ and $B$ shown in figure is :View Solution
- 4Masses of $3$ wires of same metal are in the ratio $1 : 2 : 3$ and their lengths are in the ratio $3 : 2 : 1$. The electrical resistances are in ratioView Solution
- 5If the ammeter in the given circuit reads $2\, A$, the resistance $R$ is ............ $ohm$View Solution
- 6View SolutionWe are able to obtain fairly large currents in a conductor because
- 7An ammeter $A$ of finite resistance, and a resistor $R$ are joined in series to an ideal cell $C$. $A$ potentiometer $P$ is joined in parallel to $R$. The ammeter reading is $I_0$ and the potentiometer reading is $V_0$. $P$ is now replaced by a voltmeter of finite resistance. The ammeter reading now is $I$ and the voltmeter reading is $V$.View Solution
- 8A $100\, V$ voltmeter of internal resistance $20\,k\Omega $ in series with a high resistance $R$ is connected to a $110\, V$ line. The voltmeter reads $5\, V$, the value of $R$ is ................ $k \Omega $View Solution
- 9For the circuit shown, the value of current at time ${t}=3.2\, {s}$ will be ...... ${A}$.View Solution
[Voltage distribution $V(t)$ is shown by Fig. $(1)$ and the circuit is shown in Fig. $(2)$]
- 10If $400\; \Omega$ of resistance is made by adding four $100\; \Omega$ resistance of tolerance $5 \%$ then the tolerance of the combination is .....$\%$View Solution