Question

The air temperature above coastal areas is profoundly influenced by the large specific heat of water. One reason is that the energy released when $1{\text{m}^{3}}$ of water cools by $1{\text{ }}^{\circ }$C will raise the temperature of a much larger volume of air by $1{\text{ }}^{\circ }$C. Find this volume of air. The specific heat of air is approximately $1{\text{ }}kJ/kg^{\circ }$C. Take the density of air to be $1.3{\text{ }}kg/{m^3}$.
(A) $2 \times {10^3}{m^3}$
(B) $3 \times {10^3}{m^3}$
(C) $1 \times {10^3}{m^3}$
(D) $4 \times {10^3}{m^3}$

Answer 1

Hint Use the formula for specific heat ${Q_c} = mc\Delta T$,for both the cases, for water and for air and then equate them to find the volume of air.

Complete step by step solution
We know that the density of water is: $\rho = 1kg/{m^3}$ .
From the question, we will take volume of water to be: $V = 1{m^3}$
The mass of this volume of water will be: $m = \rho V$
Substituting the values of density and volume in the expression of mass gives us the mass of water to be taken.
$\Rightarrow m = 1 \times {10^3}kg/{m^3} \times 1{m^3} \\\ \Rightarrow m = 1 \times {10^3}kg \\\$
Specific heat of water is $4186{\text{ J/kg}^{\circ }}$C.
Heat energy released by water when it cools by ${1}^{\circ }$C is, ${Q_c} = mc\Delta T$
Where, m is the mass of water
c is the specific heat of water
and $\Delta T$ is the change in temperature of water

$$\Rightarrow {Q_C} = (1 \times {10^3}kg)(4186J/kg^\circ C)( - 1^{\circ }C) \\\ \Rightarrow {Q_C} = - 4 \times {10^6}J \\\$$

When this amount of heat is released by water, and transferred to air, the temperature rises by $1^\circ C$. Let the volume of air whose temperature rises by $1{\text{ }}^\circ C$ be $V{\text{ }}{m^3}$.
Again using the formula ${Q_c} = mc\Delta T$ and substituting the value of mass of air to be $m = \rho V$ , where $\rho$is the density of air given to be $\rho = 1.3kg/{m^3}$ and the specific heat of air is approximately $1{\text{ }}kJ/kg^\circ C$, we get
$\Rightarrow {Q_C} = \rho Vc\Delta T \\\ \Rightarrow V = \dfrac{{{Q_C}}}{{\rho c\Delta T}} \\\ \Rightarrow V = \dfrac{{4 \times {{10}^6}J}}{{(1.3kg/{m^3})(1 \times {{10}^3}J/kg^\circ C)(1^\circ C)}} \\\ \Rightarrow V = 3 \times {10^3}{m^3} \\\$

Therefore, option (B) is correct.

Note The negative sign in the expression of ${Q_C}$ for water represents that the heat energy is being released and the positive sign would mean that the heat energy has been absorbed.