Chapter 22, Problem 25A

To determine

(a)

The percent increase in temperature.

Answer to Problem 25A

Percentage increase in temperature is $13\%$.

Explanation of Solution

Given information:

Initial temperature of heated steel is ${1300}^0$ F and color is dark red. Color changes to dull cherry-red at approximately ${1470}^0$.

Calculation:

As per the given information initial temperature is ${1300}^0$ F and color is dark red. Color changes to dull cherry-red at approximately ${1470}^0$.

So, change in temperature $=$$({1470}^0-{1300}^0)={170}^{0}F$.

Need to find the percentage change in the temperature.

As we know the proportion to find percentage is given as

  $\frac{P}{B}=\frac{R}{100}$   where, $P=\text{percentage, }B=\text{ base and }R=\text{rate}$

Here, initial temperature will be base $B=$${1300}^{0}F$, percentage will be $P=$${170}^{0}F$.

Putting these value in standard proportion,

  $\begin{array}{l}\frac{P}{B}=\frac{R}{100}\\ \Rightarrow \frac{170}{1300}=\frac{R}{100}\text{ {Put }B=1300\text{ and }P=170\text{}}\\ \text{Applying cross multiplication,}\\ \Rightarrow 1300R=170\times 100\\ \Rightarrow R=\frac{170\times 100}{1300}\\ \Rightarrow R=13.07692\\ \Rightarrow R=13\text{ {Rounding off to nearest whole number}}\end{array}$

Hence, percentage increase in temperature is $13\%$.

To determine

(b)

The percent increase in temperature.

Answer to Problem 25A

Percentage increase in temperature is $69\%$.

Explanation of Solution

Given information:

Initial temperature of heated steel is ${1300}^0$ F and color is dark red. Color changes to orange yellow at approximately ${2200}^0$.

Calculation:

As per the given information initial temperature is ${1300}^0$ F and color is dark red. Color changes to orange yellow at approximately ${2200}^0$.

So, change in temperature $=$$({2200}^0-{1300}^0)={900}^{0}F$.

Need to find the percentage change in the temperature.

As we know the proportion to find percentage is given as

  $\frac{P}{B}=\frac{R}{100}$   where, $P=\text{percentage, }B=\text{ base and }R=\text{rate}$

Here, initial temperature will be base $B=$${1300}^{0}F$, percentage will be $P=$${900}^{0}F$.

Putting these value in standard proportion,

  $\begin{array}{l}\frac{P}{B}=\frac{R}{100}\\ \Rightarrow \frac{900}{1300}=\frac{R}{100}\text{ {Put }B=1300\text{ and }P=900\text{}}\\ \text{Applying cross multiplication,}\\ \Rightarrow 1300R=900\times 100\\ \Rightarrow R=\frac{900\times 100}{1300}\\ \Rightarrow R=69.23076\\ \Rightarrow R=69\text{ {Rounding off to nearest whole number}}\end{array}$

Hence, percentage increase in temperature is $69\%$.

To determine

(c)

The percent increase in temperature.

Answer to Problem 25A

Percentage increase in temperature is $110\%$.

Explanation of Solution

Given information:

Initial temperature of heated steel is ${1300}^0$ F and color is dark red. Color changes to brilliant white at approximately ${2730}^0$.

Calculation:

As per the given information initial temperature is ${1300}^0$ F and color is dark red. Color changes to brilliant white at approximately ${2730}^0$.

So, change in temperature $=$$({2730}^0-{1300}^0)={1430}^{0}F$.

Need to find the percentage change in the temperature.

As we know the proportion to find percentage is given as

  $\frac{P}{B}=\frac{R}{100}$   where, $P=\text{percentage, }B=\text{ base and }R=\text{rate}$

Here, initial temperature will be base $B=$${1300}^{0}F$, percentage will be $P=$${1430}^{0}F$.

Putting these value in standard proportion,

  $\begin{array}{l}\frac{P}{B}=\frac{R}{100}\\ \Rightarrow \frac{1430}{1300}=\frac{R}{100}\text{ {Put }B=1300\text{ and }P=1430\text{}}\\ \text{Applying cross multiplication,}\\ \Rightarrow 1300R=1430\times 100\\ \Rightarrow R=\frac{1430\times 100}{1300}\\ \Rightarrow R=110\\ \Rightarrow R=110\text{ {Rounding off to nearest whole number}}\end{array}$

Hence, percentage increase in temperature is $110\%$.