Question

Water of volume $2L$ in a container is heated with a coil of $1kW$ at ${27^\circ }C$ . The lid of the container is open and energy dissipates at a rate of $160\,J.{s^{ - 1}}$ . In how much time, temperature rises from ${27^\circ }C\,\,to\,{77^\circ }C$ ? [Given specific heat of water is $4.2kJ.K{g^{ - 1}}$ ]
(A) $8\min 20s$
(B) $6\min 2s$
(C) $7\min $
(D) $14\min $

Answer 1

Hint: In order to answer this question, first we will find the net heat energy generated after the energy dissipates, and then we will find the mass of the water, as we know the density and volume of water. And then we will apply the formula of net Heat energy in terms of mass, specific heat, change in temperature and the time taken to the change in temperature.

Formula used:-
Density in terms of mass and volume: $Density = \dfrac{{mass}}{{volume}}$
Net Heat generated, $\therefore Q = m \times S \times (\Delta T)$ and $Q = E \times t$ .

Complete answer:
According to the question, after the given volume of water is heated, the power generated is: $\therefore P = 1kW = 1000W$
So, here, thermal energy takes place.
And, the energy dissipates at the rate of $160\,J.{s^{ - 1}}$ ,
So, the net heat energy is, $ = 1000 - 160 = 840J$
The heat energy that increases the temperature from ${27^\circ }C\,\,to\,{77^\circ }C$ is, ${Q_{net}} = 840J$
Now, as we know, the density of water is $1000kg.{m^{ - 3}}$ .
And, the volume of water is given, $V = 2L$
We have both density and volume of water, so we can find the mass of the water by applying the formula of density in the terms of volume and mass:
$
  \therefore Density = \dfrac{{Mass}}{{Volume}} \\
   \Rightarrow Mass = 1000 \times 2 \\
   \Rightarrow m = 2000g = 2kg \\
 $
Now, as we know that the formula of heat energy in the terms of time, specific heat and the mass of water:
$\therefore Q = m \times S \times (\Delta T)$ …….(i)
where, $Q$ is the net heat energy remains after dissipation,
$m$ is the mass of the water,
$S$ is the specific heat, which is given,
$\Delta T$ is the change in temperature.
So, the change in temperature, $\Delta T = {77^\circ }C - {27^\circ }C = {50^\circ }C$
We will also write a formula of $Q$ in the terms of time and the thermal energy:
$Q = E \times t$ ………(ii)
where, $E$ is the thermal energy, $840J$
and, $t$ is the time taken to change the temperature.
So, as comparing eq(i) and eq(ii), we get:-
$\therefore t = \dfrac{{m \times S \times \Delta T}}{E}$
Now, we will substitute the all calculated values and the given values:
$ \Rightarrow t = \dfrac{{2 \times 4200 \times 50}}{{840}} = 500\sec ond = 8\min 20\sec $
Therefore, the time taken at which the temperature rises from ${27^\circ }C\,\,to\,{77^\circ }C$ is $8\min 20s$ .
Hence, the correct option is (A) $8\min 20s$ .

Note:
Energy moves from an initial form to a final form in a dissipative process, with the final firm's capacity to accomplish mechanical work being less than the beginning form's. Heat transfer, for example, is dissipative because it involves the transfer of internal energy from a hotter to a cooler body.