Question

-. The temperature of air above coastal areas is greatly influenced by the large

specific heat of water. The specific heat of air between temperatures of 40°F and
90°F is about 1,000.0 J/kg • K. Consider the situation where 4,186 J of energy is
given up by 1.0 kg of water, causing its temperature to drop by 1.0 K.
a. If the air over the water increases temperature by 5.5 K,
what is its mass?

What is the volume of The air over the water, assuming that the airs density is 1.29 kg/m^3

Answer 1

The mass of air that would increase in temperature by 5.5 K due to the given energy of 4186 J is 0.76 kg, and its volume, assuming a density of 1.29 kg/m³, would be 0.59 m³.

Step-by-step explanation:

The calculation of the mass of air heated by the given energy involves using the specific heat capacity formula. The specific heat capacity formula is Q = mcΔT, where Q is the heat energy transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Given the specific heat capacity of air is approximately 1000 J/kg·K, and we have the amount of energy as 4186 J (from the water) causing an increase in temperature of 5.5 K for the air, we can rearrange the formula to solve for the mass m as follows:

m = Q / (cΔT)

m = 4186 J / (1000 J/kg·K × 5.5 K)

m = 4186 J / 5500 J/kg

m = 0.76 kg

The mass of air that would increase in temperature by 5.5 K due to the given energy is 0.76 kg.

To find the volume of this mass of air, we use the density relationship, which is density (p) = mass (m) / volume (V). Given the density of air is 1.29 kg/m³, we can solve for the volume V as follows:

V = m / p

V = 0.76 kg / 1.29 kg/m³

V = 0.59 m³

The volume of air over the water, assuming a density of 1.29 kg/m³, would be 0.59 m³.