Chapter 1, Problem 50P

(a)

To determine

To determine: The factor by which $\Delta t$ will get larger or smaller if $d$ is made three times larger.

Answer to Problem 50P

Answer: The $\Delta t$ will get smaller if $d$ is made three times larger by the factor $\frac{1}{9}$.

Explanation of Solution

Explanation:

Given information:

The $d$ is made three times larger.

The given equation for thermal conduction of energy in to cylindrical block of ice is,

$\frac{Q}{\Delta t}=\frac{k\pi d^2\left(T_b-T_c\right)}{4L}$

For experimental control, in one set of trials all quantities except $d$ and $\Delta t$ are constant.

For all quantities are constant except $d$ and $\Delta t$, this equation can be written as,

$\Delta t\propto \frac{1}{d^2}$

And this equation can be written as,

$\frac{\Delta t_1}{\Delta t_2}={\left(\frac{d_2}{d_1}\right)}^2$

• $\Delta t_1$ is the initial time change.
• $d_1$ is the initial diameter.
• $\Delta t_2$ is the final time change.
• $d_2$ is the final diameter.

Substitute $3d_1$ for $d_2$ to find the $\Delta t_2$ in terms of $\Delta t_1$.

$\begin{array}{c}\frac{\Delta t_1}{\Delta t_2}={\left(\frac{3d_1}{d_1}\right)}^2\\ \Delta t_2=\frac{1}{9}\Delta t_1\end{array}$

Conclusion:

Therefore, the $\Delta t$ will get smaller if $d$ is made three times larger by the factor $\frac{1}{9}$.

(b)

To determine

To determine: The pattern of proportionality of $\Delta t$ to $d$ does the equation predict.

Answer to Problem 50P

Answer: The pattern of inverse square proportionality of $\Delta t$ to $d$ does the equation predicts.

Explanation of Solution

Explanation:

Given information:

The $d$ is made three times larger.

The given equation thermal conduction of energy in to cylindrical block of ice is,

$\frac{Q}{\Delta t}=\frac{k\pi d^2\left(T_b-T_c\right)}{4L}$

For experimental control, in one set of trials all quantities except $d$ and $\Delta t$ are constant.

For all quantities are constant except $d$ and $\Delta t$, this equation can be written as,

$\Delta t\propto \frac{1}{d^2}$

Its means $\Delta t$ is inversely proportional to square of $d$.

Conclusion:

Therefore, the pattern of inverse square proportionality of $\Delta t$ to $d$ does the equation predicts.

(c)

To determine

To determine: To display this proportionality as a straight line on a graph, the quantities should be plot on the horizontal and vertical axis.

Answer to Problem 50P

Answer: To display this proportionality as a straight line on a graph, the quantities should be plot on the horizontal axis is $\Delta t$ and $\frac{1}{d^2}$ on the vertical axis.

Explanation of Solution

Explanation:

Given information:

The given equation thermal conduction of energy in to cylindrical block of ice is,

$\frac{Q}{\Delta t}=\frac{k\pi d^2\left(T_b-T_c\right)}{4L}$

For experimental control, in one set of trials all quantities except $d$ and $\Delta t$ are constant.

To display this proportionality as a straight line on a graph, the quantities should be plot on the horizontal axis is $\Delta t$ because it is an independent variable and other i.e. $\frac{1}{d^2}$ on the vertical axis or $y$-axis because it is a dependent variable.

Conclusion:

Therefore, to display this proportionality as a straight line on a graph, the quantities should be plot on the horizontal axis is $\Delta t$ and $\frac{1}{d^2}$ on the vertical axis.

(d)

To determine

To determine: The expression represents the theoretical slope of the graph.

Answer to Problem 50P

Answer: The expression represents the theoretical slope of the graph is $\frac{k\pi \left(T_b-T_c\right)}{4LQ}$.

Explanation of Solution

Explanation:

Given information:

The given equation thermal conduction of energy in to cylindrical block of ice is,

$\frac{Q}{\Delta t}=\frac{k\pi d^2\left(T_b-T_c\right)}{4L}$

For experimental control, in one set of trials all quantities except $d$ and $\Delta t$ are constant.

To calculate the slope of this equation is,

$\frac{1}{d^2}=\frac{k\pi \left(T_b-T_c\right)}{4LQ}\Delta t$

The general equation of straight line $y=mx$ is compared to the above equation,

$m=\frac{k\pi \left(T_b-T_c\right)}{4LQ}$

Conclusion:

Therefore, the expression represents the theoretical slope of the graph is $\frac{k\pi \left(T_b-T_c\right)}{4LQ}$.