Chapter 24, Problem 1P

Perform the same computation as Sec. 24.1, but compute the amount of heat required to raise the temperature of 1200 g of the material from $-150\text{ to }100°\text{C}$. Use Simpson's rule for your computation, with values of $T\text{ at 50}°\text{C}$ increments.

To determine

To calculate: The amount of heat required of $1200\text{ g}$ material by using Simpson’s rule, when temperature is increased from $-150°\text{C}$ to $100°\text{C}$.

Answer to Problem 1P

Solution:

The amount of heat required of $1200\text{ g}$ material by using Simpson’s ruleis $38892\text{ cal}$.

Explanation of Solution

Given Information:

Write the value of the provided temperatures.

$T_1=-150\text{°C}$ and $T_2=100\text{°C}$

Write the value of the provided mass of the material.

$m=1200\text{ g}$

Write the value of the temperature increments.

$\Delta T=50\text{°C}$

Formula used:

Write the expression of heat required when temperature is increased.

$\Delta H=m{\displaystyle {\int }_{T_1}^{T_2}c\left(T\right)dT}$ ,

And,

$\Delta T=T_2-T_1$

Write the expression of the heat capacity of the material, when temperature is increased.

$c\left(T\right)=0.132+1.56\times {10}^{-4}T+2.64\times {10}^{-7}{T}^2$

The formula of the 1/3 Simpson’s rule for the number integration is as follows:

${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\frac{h}{6}\left(f\left(a\right)+4f\left(\frac{a+b}{2}\right)+f\left(b\right)\right)$

The formula of the 3/8 Simpson’s rule for the number integration is as follows:

${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\frac{h}{8}\left(f\left(a\right)+3f\left(\frac{2a+b}{3}\right)+3f\left(\frac{a+2b}{3}\right)+f\left(b\right)\right)$

Here, $h=b-a$

Calculation:

Calculate the value of the of $c\left(T\right)$ at different values of $T$.

For $T=-150\text{°C}$,

$\begin{array}{c}c\left(T=-150\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(-150\text{°C}\right)+2.64\times {10}^{-7}{\left(-150\text{°C}\right)}^2\\ =0.11454\text{ cal/g}\cdot \text{°C}\end{array}$

For $T=-100\text{°C}$,

$\begin{array}{c}c\left(T=-100\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(-100\text{°C}\right)+2.64\times {10}^{-7}{\left(-100\text{°C}\right)}^2\\ =0.11904\text{ cal/g}\cdot \text{°C}\end{array}$

For $T=-50\text{°C}$,

$\begin{array}{c}c\left(T=-50\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(-50\text{°C}\right)+2.64\times {10}^{-7}{\left(-50\text{°C}\right)}^2\\ =0.12486\text{ cal/g}\cdot \text{°C}\end{array}$

For $T=0\text{°C}$,

$\begin{array}{c}c\left(T=0\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(0\text{°C}\right)+2.64\times {10}^{-7}{\left(0\text{°C}\right)}^2\\ =0.132\text{ cal/g}\cdot \text{°C}\end{array}$

For $T=50\text{°C}$,

$\begin{array}{c}c\left(T=50\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(50\text{°C}\right)+2.64\times {10}^{-7}{\left(50\text{°C}\right)}^2\\ =0.14046\text{ cal/g}\cdot \text{°C}\end{array}$

For $T=100\text{°C}$,

$\begin{array}{c}c\left(T=100\text{°C}\right)=0.132+1.56\times {10}^{-4}\left(100\text{°C}\right)+2.64\times {10}^{-7}{\left(100\text{°C}\right)}^2\\ =0.15024\text{ cal/g}\cdot \text{°C}\end{array}$

Construct the table for heat capacity $c\left(T\right)$ for different value of $T$.

$\begin{array}{cc}T\left(\text{°C}\right)& c\left(T\right)\left(\text{cal/g}\cdot \text{°C}\right)\\ -150& 0.11454\\ -100& 0.11904\\ -50& 0.12486\\ 0& 0.132\\ 50& 0.14046\\ 100& 0.15024\end{array}$

Calculate the amount of heat required to increase the temperature of 1200g of material from $T_1=-150\text{°C}$ to $T_2=100\text{°C}$.

$\Delta H=m{\displaystyle {\int }_{-150}^{100}c\left(T\right)dT}$ …… (1)

Use the integral in equation (1) by dividing the limits in two parts, that can be written as,

$\Delta H=m\left({\displaystyle {\int }_{-150}^{-50}c\left(T\right)dT}+{\displaystyle {\int }_{-50}^{100}c\left(T\right)dT}\right)$ …… (2)

Consider the integral as follows:

$I={\displaystyle {\int }_{-150}^{-50}c\left(T\right)dT}+{\displaystyle {\int }_{-50}^{100}c\left(T\right)dT}$ …… (3)

Consider first part of the integral of $I$ is,

$I_1={\displaystyle {\int }_{-150}^{-50}c\left(T\right)dT}$

Solve the above integral by using Simpson’s 1/3^rd rule.

$\begin{array}{c}{I}_1=\frac{-50-\left(-150\right)}{6}\left(f\left(-150\right)+4f\left(\frac{-150+\left(-50\right)}{2}\right)+f\left(-50\right)\right)\\ =\frac{100}{6}\left(f\left(-150\right)+4f\left(-100\right)+f\left(-50\right)\right)\\ =\frac{100}{6}\left(0.11454+4\left(0.11904\right)+0.12486\right)\\ =11.926\text{ cal/g}\end{array}$

Consider the second part of integral $I$ is,

$I_2={\displaystyle {\int }_{-50}^{100}c\left(T\right)dT}$

Solve the above integral by using Simpson’s 3/8^th rule.

$\begin{array}{c}{I}_2=\frac{100-\left(-50\right)}{8}\left(f\left(-50\right)+3f\left(\frac{2\left(-50\right)+100}{3}\right)+3f\left(\frac{-50+2\left(100\right)}{3}\right)+f\left(100\right)\right)\\ =\frac{150}{8}\left(f\left(-50\right)+3f\left(0\right)+3f\left(50\right)+f\left(100\right)\right)\\ =\frac{150}{8}\left(0.12486+3\left(0.132\right)+3\left(0.14046\right)+0.15024\right)\\ =20.484\text{ cal/g}\end{array}$

Substitute $20.484\text{ cal/g}$ for $I_2$ and $11.926\text{ cal/g}$ for $I_1$ in equation (3).

$\begin{array}{c}I=I_1+I_2\\ =11.926\text{ cal/g}+20.484\text{ cal/g}\\ \text{=}32.41\text{ cal/g}\end{array}$

Substitute $32.41\text{ cal/g}$ for $I$ and $1200\text{ g}$ for $m$ in equation (2).

$\begin{array}{c}\Delta H=mI\\ =\left(1200\text{ g}\right)\left(32.41\text{ cal/g}\right)\\ =38892\text{ cal}\end{array}$

Therefore, the heat required to increase temperature of the material by using Simpson’s ruleis $38892\text{ cal}$.