Question

A step supplies water at $22^{\circ} \mathrm{C}$. A man takes 1 liter per minute at $37^{\circ} \mathrm{C}$ from the geyser. the power of geyser is
A.$2100~\text{W}$
B. $1050 \mathrm{~W}$
C. 1575 w
D. $525 \mathrm{~W}$

Answer 1

Hint: Electrical power is the rate per unit of time at which an electrical circuit transmits electrical energy. A watt, one joule per second, is the SI unit of power. Calculate Heat required by water in per second in $\dfrac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=250 \mathrm{cal} \mathrm{s}^{-1}$ and then change it to SI unit of the power $\dfrac{\mathrm{dQ}}{\mathrm{dt}}=250 \times 4.2 \mathrm{Js}^{-1}$

Formula used:
$\dfrac{\mathrm{dQ}}{\mathrm{dt}}=\dfrac{\mathrm{dm}}{\mathrm{dt}} \mathrm{C}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$

Complete solution:
Electrical Power, (P) is the rate at which energy is absorbed or generated within a circuit in a circuit. While the connected load absorbs it, a source of energy such as a voltage will produce or deliver power. Electrical power is the rate per unit of time at which an electrical circuit transmits electrical energy. A watt, one joule per second, is the SI unit of power. Generally, electricity is generated by electric generators but can also be supplied by sources such as electric batteries.

Given
$\mathrm{T}_{1}=22^{\circ} \mathrm{C} \quad \mathrm{T}_{2}=37^{\circ} \mathrm{C} \quad \mathrm{m}=1$ liter $\text{t}=1~\text{min}$

Heat required by water per second to change its temperature from $22^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$ $\dfrac{\mathrm{dQ}}{\mathrm{dt}}=\dfrac{\mathrm{dm}}{\mathrm{dt}} \mathrm{C}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$
$\dfrac{\mathrm{dQ}}{\mathrm{dt}}=\dfrac{1000}{60}(1)(37-22)$
$\dfrac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=250 \mathrm{cal} \mathrm{s}^{-1}$
$\dfrac{\mathrm{dQ}}{\mathrm{dt}}=250 \times 4.2 \mathrm{Js}^{-1}$
$\dfrac{\mathrm{dQ}}{\mathrm{dt}}=1050 \mathrm{Js}^{-1}$

The power of geyser is $\dfrac{\mathrm{dQ}}{\mathrm{dt}}=1050 \mathrm{Js}^{-1}$. Correct option is (B).

Note:
According to \[P=IV\], a circuit element dissipates or generates power, where I is the current through the element and V is the voltage across it. The instantaneous power \[p\left( t \right)=i\left( t \right)v\left( t \right)\]is also time-dependent, since both the current and the voltage depend on time in an ac circuit. These three kinds of power, true, reactive, and apparent, relate in trigonometric form to each other.