26.2 Convection  (Page 2/4)

Strategy

Heat is used to raise the temperature of air so that $Q=\text{mc}\text{Δ}T$ . The rate of heat transfer is then $Q/t$ , where $t$ is the time for air turnover. We are given that $\text{Δ}T$ is $\text{10}\text{.}\mathrm{0º}\text{C}$ , but we must still find values for the mass of air and its specific heat before we can calculate $Q$ . The specific heat of air is a weighted average of the specific heats of nitrogen and oxygen, which gives $c=c_{\text{p}}\cong \text{1000}\text{ J/kg}\cdot º\text{C}$ from [link] (note that the specific heat at constant pressure must be used for this process).

Solution

1. Determine the mass of air from its density and the given volume of the house. The density is given from the density $\rho$ and the volume
   $m=\text{ρV}=\left(1\text{.}\text{29}{\text{kg/m}}^3\right)\left(\text{12}\text{.}0\text{m}\times \text{18}\text{.}0\text{m}\times 3.00\text{m}\right)=\text{836}\text{kg.}$
2. Calculate the heat transferred from the change in air temperature: $Q=\text{mc}\text{Δ}T$ so that
   $Q=\left(\text{836}\text{kg}\right)\left(\text{1000 J/kg}\cdot º\text{C}\right)\left(\text{10.0º}\text{C}\right)\text{= 8}\text{.}\text{36}\times {\text{10}}^6\text{J.}$
3. Calculate the heat transfer from the heat $Q$ and the turnover time $t$ . Since air is turned over in $t=0\text{.}500\text{h}=\text{1800}\text{s}$ , the heat transferred per unit time is
   $\frac{Q}{t}=\frac{8\text{.}\text{36}\times {\text{10}}^{\mathrm{6 }}\text{J}}{\text{1800 }\text{s}}=4\text{.}64\text{kW.}$

Discussion

This rate of heat transfer is equal to the power consumed by about forty-six 100-W light bulbs. Newly constructed homes are designed for a turnover time of 2 hours or more, rather than 30 minutes for the house of this example. Weather stripping, caulking, and improved window seals are commonly employed. More extreme measures are sometimes taken in very cold (or hot) climates to achieve a tight standard of more than 6 hours for one air turnover. Still longer turnover times are unhealthy, because a minimum amount of fresh air is necessary to supply oxygen for breathing and to dilute household pollutants. The term used for the process by which outside air leaks into the house from cracks around windows, doors, and the foundation is called “air infiltration.”

A cold wind is much more chilling than still cold air, because convection combines with conduction in the body to increase the rate at which energy is transferred away from the body. The table below gives approximate wind-chill factors, which are the temperatures of still air that produce the same rate of cooling as air of a given temperature and speed. Wind-chill factors are a dramatic reminder of convection’s ability to transfer heat faster than conduction. For example, a 15.0 m/s wind at $\mathrm{0º}\text{C}$ has the chilling equivalent of still air at about $-\text{18º}\text{C}$ .

Although air can transfer heat rapidly by convection, it is a poor conductor and thus a good insulator. The amount of available space for airflow determines whether air acts as an insulator or conductor. The space between the inside and outside walls of a house, for example, is about 9 cm (3.5 in) —large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. Similarly, the gap between the two panes of a double-paned window is about 1 cm, which prevents convection and takes advantage of air’s low conductivity to prevent greater loss. Fur, fiber, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection, as shown in the figure. Fur and feathers are lightweight and thus ideal for the protection of animals.