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^@C` will raise the temperature of a much larger volume of air by `1^@C`. Find this volume of air. The specific heat of air is approximately `1 kJ//kg^@C`. Take the density of air to be `1.3 kg//m^3`.

Explanation

The mass of `1m^3` of water is specified by its density,
`m=rhoV=(1.00xx10^3 kg//m^(3)))(1m^3)=1xx10^3kg`
When `1m^3` of water cools by `1^@C`, it releases energy
`Q_c=mcDeltaT=(1xx10^3kg)4186 J//kg.^@C(-1^@C)=-4xx10^6J`
Where the negative sign represents heat output. When `+4xx10^6J` is transferred to the air, raising its temperature by `1^@C`, the volume of the air is given by `Q_c=mcDeltaT=rhoVcDeltaT`
`V=(Q_c)/(rhocDeltaT)=(4xx10^6J)/((1.3kg//m^3)(1xx10^3J//kg^@C)(1^@C))=3xx10^3m^2`
The volume of the air is a thousand times larger than the volume of the water.