Question

Water at temperature 20°C flows from a tap T into a container C at the rate of 0.021 litre per minute. The heating element (resister R)
generates sufficient power P so that water in the container boils and gets converted into steam. At steady state, it is found that the amount
of water in the container neither increases nor decreases with time. Neglecting heat losses, the value of P(in W) is given by
(to the nearest integer)
[For water specific heat capacity = 4.2 kJ/kg K; latent heat of vaporization = 2.3 MJ/kg]
100
T
3
020​

Answer 1

Answer:

Please Give A Relevant Question

Answer 2

Answer:

Neglecting the losses the power(P) in the container is $922.6W.$

Explanation:

The rate of water(R) coming out from tap T is 0.021 liter per minute

i.e.$0.021\times 10^{-3} m^{3} /min$

R=$\frac{2.1\times10^{-5} }{60}$ $m^{3} /s$

The same amount of water should be ejected in the form of steam.

(As we are neglecting the heat losses)

So heat supplied will be,

$\Delta$$Q=$$ms$$\Delta$$T+$$mL$                (1)

Where,

$m=$mass of the water

$s=$ specific heat capacity

$\Delta$$T=$ change in temperature

$L=$latent heat of vapourization

We can find the rate of heat transfer by dividing equation (1) with time(t) i.e.

$\frac{\Delta Q}{t}=\frac{ms \Delta T}{t}+\frac{mL}{t}$      (2)

And some of the values given in the question are;

$s=4.2kJ/kg-K$

$L=2.3 MJ/kg=2.3$×$10^{3} kJ/kg$$=2300kJ/kg$

We know that rate of change of heat is power(P).

So, $\Delta$$Q/t$ can be written as $P$.

$\frac{m}{t}$$=$$\frac{\rho\times V}{t}$$=$$\rho\times R$

So, equation (2) can be written as;

$P=$$\rho\times$$Rs$$\Delta$$T+$$\rho$$R$$L$       (3)

$\Delta$$T=100\textdegree C$$-20\textdegree C$ $=80\textdegree C$   

(We have taken 100°C as the water in the container boils)

By substituting the values in equation (3) we get;

$P=\frac{1000\times2.1\times10^{-5}\times4.2\times80\textdegree}{60} +$$\frac{1000\times2.1\times10^{-5}\times2300}{60} kJ/Sec$

$P=922.6 J/sec\\P=922.6 W$

Hence, neglecting the losses the power(P) in the container is $922.6W$.