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 $1m^3$ of water cools by $1^{\circ }C$ will raise the temperature of a much larger volume of air by $1^{\circ }C$. Find this volume of air. The specific heat of air is approximately $1kJ/k{g}^{\circ }C$. Take the density of air to be $1.3kg/m^3$.

• 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

The correct option is B $3\times {10}^3{m}^3$

The mass of $1m^3$ of water is specified by its density,
$m=\rho V=(1.00\times {10}^{3}kg/m^3)\left(1m^3\right)=1\times {10}^{3}kg$

When $1m^3$ of water cools by $1^{\circ }C$, it releases energy
$Q_c=mc\Delta T=(1\times {10}^{3}kg)\left(4186J/k{g}^{\circ }C\right)(-1^{\circ }C)=-4\times {10}^{6}J$

Where the negative sign represents heat output. When $+4\times {10}^{6}J$ is transferred to the air, raising its temperature by $1^{\circ }C$, the volume of the air is given by $Q_c=mc\Delta T=\rho Vc\Delta T$
$V=\frac{Q_c}{\rho c\Delta T}=\frac{4\times {10}^{6}J}{1.3kg/m^3(1\times {10}^{3}J/k{g}^{\circ }C)\left(1^{\circ }C\right)}=3\times {10}^3{m}^3$