Question

On a winter day the temperature of the tap water is $20^\circ {\text{C}}$ whereas the room temperature is $5^\circ {\text{C}}$. Water is stored in a tank of capacity $0.5\,{{\text{m}}^3}$ for household use. If it were possible to use the heat liberated by the water to lift a $10\,{\text{kg}}$ mass vertically, how high can it be lifted as the water comes to the room temperature? Take $g = 10\,{\text{m/}}{{\text{s}}^2}$.
A. $215\,{\text{km}}$
B. $315\,{\text{km}}$
C. $115\,{\text{km}}$
D. $105\,{\text{km}}$

Answer 1

Determine the mass of the water in the tank using the formula for density of the water in terms of mass and volume of the water. Use the formula for energy released by the water when its temperature is decreased to room temperature and the formula for potential energy required to lift the mass vertically.

Formulae used:
The amount of energy $E$ released by a material is
$E = mC\Delta T$ …… (1)
Here, $m$ is the mass of the material, $C$ is the specific heat of the material and $\Delta T$ is the change in temperature of the material.
The density $\rho$ of an object is
$\rho = \dfrac{m}{V}$ …… (2)
Here, $m$ is the mass of the object and $V$ is the volume of the object.
The potential energy $U$ of an object is
$U = mgh$ …… (3)
Here, $$$$ is the mass of the object, $g$ is acceleration due to gravity and $h$ is the height of the object from the ground.

Complete step by step answer:
We have given that the temperature of the water coming out of the tap is $20^\circ {\text{C}}$ and the room temperature is $5^\circ {\text{C}}$.
The volume of the water in the tank is $0.5\,{{\text{m}}^3}$.
$V = 0.5\,{{\text{m}}^3}$

The mass that we want to lift by the energy liberated by water is $10\,{\text{kg}}$.
$m = 10\,{\text{kg}}$

Let us first determine the mass ${m_w}$ of the water in the tank using equation (2).
Substitute $1000\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$ and $0.5\,{{\text{m}}^3}$ for $V$ in equation (2).
$1000\,{\text{kg/}}{{\text{m}}^3} = \dfrac{{{m_w}}}{{0.5\,{{\text{m}}^3}}}$
$\Rightarrow {m_w} = 500{\text{kg}}$
Hence, the mass of water in the tank is $500{\text{kg}}$.

The specific heat of the water is approximately ${\text{4200}}\,{\text{J}}$.
$C = {\text{4200}}\,{\text{J}}$

The change in temperature of the water when it comes out of the tap is
$\Delta T = {\text{20}}^\circ {\text{C}} - {\text{5}}^\circ {\text{C}}$
$\Rightarrow \Delta T = 15^\circ {\text{C}}$

The energy $E$ liberated by the water when its temperature is decreased must be equal to the potential energy $U$ required to life the mass vertically.
$E = U$

Substitute ${m_w}C\Delta T$ for $E$ and $mgh$ for $U$ in the above equation.
${m_w}C\Delta T = mgh$
$\Rightarrow h = \dfrac{{{m_w}C\Delta T}}{{mg}}$

Substitute $500{\text{kg}}$ for ${m_w}$, ${\text{4200}}\,{\text{J}}$ for $C$, $15^\circ {\text{C}}$ for $\Delta T$, $10\,{\text{kg}}$ for $m$ and $10\,{\text{m/}}{{\text{s}}^2}$ for $g$ in the above equation.
$\Rightarrow h = \dfrac{{\left( {500{\text{kg}}} \right)\left( {{\text{4200}}\,{\text{J}}} \right)\left( {15^\circ {\text{C}}} \right)}}{{\left( {10\,{\text{kg}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)}}$
$\Rightarrow h = 315000\,{\text{m}}$
$\Rightarrow h = \left( {315000\,{\text{m}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{km}}}}{{1\,{\text{m}}}}} \right)$
$\therefore h = 315\,{\text{km}}$

Therefore, the water can be lifted to the height of $315\,{\text{km}}$. Hence, the correct option is B.

Note: The actual specific heat for water is ${\text{4185}}\,{\text{J}}$ but here we have rounded the value of specific heat of the water to ${\text{4200}}\,{\text{J}}$ get the answer from the given options. Also, the students should not get confused between the mass of water in the tank and mass that is to be lifted vertically.