Solveeit Logo
Login

Question

Question: A step supplies water at \(22^{\circ} \mathrm{C}\). A man takes 1 liter per minute at \(37^{\circ} \...

A step supplies water at 22∘C22^{\circ} \mathrm{C}. A man takes 1 liter per minute at 37∘C37^{\circ} \mathrm{C} from the geyser. the power of geyser is
A.2100 W2100~\text{W}
B. 1050 W1050 \mathrm{~W}
C. 1575 w
D. 525 W525 \mathrm{~W}

Explanation

Solution

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 dQdt=250cals−1\dfrac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=250 \mathrm{cal} \mathrm{s}^{-1} and then change it to SI unit of the power dQdt=250×4.2Js−1\dfrac{\mathrm{dQ}}{\mathrm{dt}}=250 \times 4.2 \mathrm{Js}^{-1}

Formula used:
dQdt=dmdtC(T2−T1)\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
T1=22∘CT2=37∘Cm=1\mathrm{T}_{1}=22^{\circ} \mathrm{C} \quad \mathrm{T}_{2}=37^{\circ} \mathrm{C} \quad \mathrm{m}=1 liter t=1 min\text{t}=1~\text{min}

Heat required by water per second to change its temperature from 22∘C22^{\circ} \mathrm{C} to 37∘C37^{\circ} \mathrm{C} dQdt=dmdtC(T2−T1)\dfrac{\mathrm{dQ}}{\mathrm{dt}}=\dfrac{\mathrm{dm}}{\mathrm{dt}} \mathrm{C}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)
dQdt=100060(1)(37−22)\dfrac{\mathrm{dQ}}{\mathrm{dt}}=\dfrac{1000}{60}(1)(37-22)
dQdt=250cals−1\dfrac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=250 \mathrm{cal} \mathrm{s}^{-1}
dQdt=250×4.2Js−1\dfrac{\mathrm{dQ}}{\mathrm{dt}}=250 \times 4.2 \mathrm{Js}^{-1}
dQdt=1050Js−1\dfrac{\mathrm{dQ}}{\mathrm{dt}}=1050 \mathrm{Js}^{-1}

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

Note:
According to P=IVP=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(t)=i(t)v(t)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.