Chapter 12.8, Problem 11P

To determine

To calculate: The heat lost through a windowpane varies jointly as the difference of the inside and outside temperatures and the window area and inversely as the thickness of the pane.

Answer to Problem 11P

Thus, the area of pane: $\approx 18$ joules.

Explanation of Solution

Given information:

The heat lost through a windowpane varies jointly as the difference of the inside and outside

temperatures and the window area and inversely as the thickness of the pane.

The heat are lost through a pane in one hour is $49.5$ joules and $40$ cm by $28$ cm that is $0.8$ cm thick when the temperature differences is ${44}^{\circ }$ C. how many joules are lost in one hour through a pane $0.5$ thick having an area that is $0.25$ times the area of the another pane when the temperature difference is ${40}^{\circ }$ C.

Formula used:

For varies directly and inversely:

  $H=K\frac{TA}{d}$

Calculation:

Consider that The heat are lost through a pane in one hour is $49.5$ joules and $40$ cm by $28$ cm that is $0.8$ cm thick when the temperature differences is ${44}^{\circ }$ C. how many joules are lost in one hour through a pane $0.5$ thick having an area that is $0.25$ times the area of the another pane when the temperature difference is ${40}^{\circ }$ C.

Recall that varies directly and inversely:

  $H=K\frac{TA}{d}$

The values are:

  $\begin{array}{l}H=49.5,T=44,A=40\text{cm}\\ \text{to}\\ A=28\text{cm}\\ \text{A=0}\text{.4}\times \text{0}\text{.28}\end{array}$

Therefore, the value of $k$ is,

  $\begin{array}{c}H=K\frac{TA}{d}\\ 49.5=K\times \frac{44\times \left(0.4\times 0.28\right)}{\frac{0.8}{100}}\\ 49.5=K\times \frac{44\times 0.112}{0.008}\\ 49.5=K\times \frac{4.928}{0.008}\end{array}$

So simplified further the value of $k$ is,

  $\begin{array}{c}49.5=K\times \frac{4.928}{0.008}\\ 49.5=K\times 616\\ \frac{49.5}{616}=K\\ 0.0803=K\end{array}$

Now to find the value of $r$ ,

  $H_2=K\frac{T_2{A}_2}{d_2}$

The values are:

  $\begin{array}{l}T=40,K=0.0803,A=0.4\times 0.28\\ d=0.5\end{array}$

Hence, when:

  $\begin{array}{c}{H}_2=K\frac{T_2{A}_2}{d_2}\\ H_2=0.0803\times \frac{40\times \frac{1}{4}\left(0.4\times 0.28\right)}{\frac{0.5}{100}}\\ H_2=0.0803\times \frac{40\times \frac{1}{4}\times 0.112}{0.005}\\ H_2=0.0803\times \frac{40\times 0.028}{0.005}\end{array}$

So,

  $\begin{array}{l}{H}_2=0.0803\times \frac{40\times 0.028}{0.005}\\ H_2=0.0803\times 224\\ H_2=17.9872\end{array}$

Thus, the area of pane: $\approx 18$ joules.