Question

Water of volume 2L, in a closed container is heated with a coil of 1kW. While water is heated, the container loses energy at 160J/s. In how much time will the temperature of water rise from ${{27}^{0}}C\text{ to 7}{{7}^{0}}C$? (specific heat of water is 4.2kJ/kg and that of container is negligible)
a)8min 20s
b)6min 2s
c)7min
d)14min

Answer 1

It is given in the question that the water is heated with a coil which supplies energy at the rate of 1000J/s. It is also given to us that the container losses heat at the rate of 160J/s. Hence whatever heat remains is used to increase the temperature of the of the water. First we will determine the amount of heat that is required to raise the temperature of water from ${{27}^{0}}C\text{ to 7}{{7}^{0}}C$. Since we now know at what rate the heat is supplied from the coil to the water, accordingly we will obtain the time taken for the process.
Formula used:
$Q=mc\Delta T$
$d=\dfrac{m}{V}$

Complete answer:
In the question it is given to us that the coil has a power of 1kW. Hence we can say that the coil supplies heat to the water at the rate of 1000J/s. It is given to us that some of the heat supplied is lost to the container at the rate of 160J/s. Therefore available heat (H)supplied per second, which is responsible for the rise in temperature of the water is equal to
$\begin{aligned} & H=1000-160 \\\ & \Rightarrow H=840J/s \\\ \end{aligned}$
Let us say a body absorbs heat Q of mass m and specific capacity c, such that the change in temperature of the body is given by $\Delta T$. Hence the amount of heat Q is equal to,
$Q=mc\Delta T$
In the above question it is given to us that the container contains 2L of water. Hence the volume in terms of cubic meter is 0.002 cubic meter. The density (d)of a substance is defined as the ratio of mass by its volume(V). Density of water is equal to $1000kg{{m}^{-3}}$. Hence the mass of the water in the container is,
$\begin{aligned} & d=\dfrac{m}{V} \\\ & \Rightarrow 1000=\dfrac{m}{2\times {{10}^{-3}}} \\\ & \Rightarrow m=2kg \\\ \end{aligned}$
Hence the amount of heat absorbed by water when its temperature changes from ${{27}^{0}}C\text{ to 7}{{7}^{0}}C$ is equal to,
$\begin{aligned} & Q=mc\Delta T \\\ & \Rightarrow Q=2kg\times 4.2\times {{10}^{3}}J/kg(77-27) \\\ & \Rightarrow Q=2\times 4.2\times {{10}^{3}}\times 50J \\\ & \Rightarrow Q=4.2\times {{10}^{5}}J \\\ \end{aligned}$
We have found that the rate at which heat is supplied to the water is H=840J/s. The total heat absorbed by the water is $4.2\times {{10}^{5}}J$ .Hence the time taken (t)for the water to change its temperature to 77 degree Celsius is,
$t=\dfrac{Q}{H}=\dfrac{4.2\times {{10}^{5}}J}{840J/s}=500s=8\min 20\sec$

Hence the correct answer of the above question is option a.

Note:
One minute has 60 sec. Hence we get 480 sec equal to 8min. In the above question it is given to us that the container is closed. Therefore there is no heat lost to the surrounding. But in reality, there is always some amount of heat which gets dissipated to the surrounding.