Q. 4.13 - An Introduction to Thermal Physics

Question

Under many conditions, the rate at which heat enters an air conditioned building on a hot summer day is proportional to the difference in temperature between inside and outside, $T_h-T_c$. (If the heat enters entirely by conduction, this statement will certainly be true. Radiation from direct sunlight would be an exception.) Show that, under these conditions, the cost of air conditioning should be roughly proportional to the square of the temperature difference. Discuss the implications, giving a numerical example.

Step-by-Step Solution

Answer

It is verified that the cost of air conditioning is approximately proportional to the square of temperature difference and the input work has to be increased by $27\%$ lowering the temperature by only $1 \mathrm{K}$ which can prove to be an economical issue sometimes.

Step 1 : Given Information

The rate of heat entering an air conditioned building on a hot day is proportional to the difference in temperature between inside and outside.

Step 2 : Formula used

Let us write the expression of input work for air conditioner

$W=Q_{\text{hot }}-Q_{\text{cold }}\dots .\text{ (1) }$

Here, $Q_{\text{hot }}$ is heat transferred to outside and $Q_{\text{cold }}$ is heat leakage from outside into the air conditioned room.

Write the expression of $Q_{\text{cold }}$

$Q_{\text{cold }}=A\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right)$

Here, $A$ is a constant, $\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right)$ is temperature difference between the outside and inside of the air conditioned room.

Write the expression of $Q_{\text{hot }}$ from the second law of thermodynamics

$Q_{\text{hot }}=Q_{\text{cold }}\left(\frac{T_{\mathrm{h}}}{T_{\mathrm{c}}}\right)$

Step 3 : Calculation

Let us substitute $Q_{\text{cold }}\left(\frac{T_{\mathrm{h}}}{T_{\mathrm{c}}}\right)$ for $Q_{\text{hot }}$ and $A\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right.)$  for $Q_{\text{cold }}$ in equation (1)

$W=\left(A\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right)\right)\left(\frac{T_{\mathrm{h}}}{T_{\mathrm{c}}}\right)-\left(A\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right)\right)$

Simplify the above expression

$W=\frac{A}{T_{\mathrm{c}}}{\left(T_{\mathrm{h}}-T_{\mathrm{c}}\right)}^2\dots ..\left(2\right)$

Assume the inside temperature is $295 \mathrm{K}$ and the outside temperature is $303 \mathrm{K}$.

Substitute $303 \mathrm{K}$ for $T_{\mathrm{h}}$ and $295 \mathrm{K}$ for $T_{\mathrm{c}}$ in equation (2)

$W_1=\frac{A}{(295 \mathrm{K})}(303 \mathrm{K}-295 \mathrm{K}{)}^2$

Assume now that the inside temperature is further lowered by $1 \mathrm{K}$. Substitute $303 \mathrm{K}$ for $T_{\mathrm{h}}$ and $294 \mathrm{K}$ for $T_{\mathrm{c}}$ in equation (2)

$W_2=\frac{A}{(294 \mathrm{K})}(303 \mathrm{K}-294 \mathrm{K}{)}^2$

The relatively extra work required to decrease the temperature by $1 \mathrm{K}$ is

$\frac{W_2}{W_1}=\frac{(295 \mathrm{K})(303 \mathrm{K}-294 \mathrm{K}{)}^2}{(294 \mathrm{K})(303 \mathrm{K}-295 \mathrm{K}{)}^2}$

$=1.27$

Step 4 : Conclusion

Thus, it is verified that the cost of air conditioning is approximately proportional to the square of temperature difference and the input work has to be increased by $27\%$ lowering the temperature by only $1 \mathrm{K}$ which can prove to be an economical issue sometimes.