Q. 5 - An Introduction to General, Organic, and Biological Chemistry

Question

In one box of nails, there are $75$ iron nails weighing $0.250lb$. The density of iron is $7.86 g/c{m}^3$. The specific heat of iron is $0.425 J/g^{\circ }C$. The melting point of iron is ${1535}^{\circ }C$..$(2.5,2.6,2.7,3.4,3.5)$

$\left(a\right)$What is the volume, in cubic centimeters, of the iron nails in the box?

$\left(b\right)$ If 30 nails are added to a graduated cylinder containing $17.6 mL$ of water, what is the new level of water, in milliliters, in the cylinder?

$\left(c\right)$How much heat, in joules, must be added to the nails in the box to raise their temperature from ${16}^{\circ }C$ to ${125}^{\circ }C$ ?

$\left(d\right)$How much heat, in joules, is required to heat one nail from ${25}^{\circ }C$ to its melting point?

Step-by-Step Solution

Answer

Part $\left(a\right)$

$\left(a\right)$The volume of iron as $V=14c{m}^3.$

Part $\left(b\right)$

$\left(b\right)$The new water level is $V=23.4 mL.$

Part $\left(c\right)$

$\left(c\right)$The heat inside the nail box is $\mathrm{Hea}t=5.6\times {10}^3\mathrm{J}.$

Part $\left(d\right)$

$\left(d\right)$the total heat in single nail is $\mathrm{Tota}l \mathrm{heat} =1400\mathrm{J}.$

Step: 1 Given information: (Part a)

Given are: $75$ iron nails and weight  $0.250lb$ and density $7.86 g/c{m}^3$ and melting point is ${1535}^{\circ }C$ and specific heatnis $0.425 J/g^{\circ }C.$

To find the volume of iron nails.

Step: 2 Equating: (Part a)

The volume of the nails may be calculated using the volume and toughness of the iron filings. The density is the mass-to-volume ratio.. 

The density as,

$\text{ Density }=\frac{\text{ mass }}{\text{ volume }}$

Rearranging,

$\mathrm{volume}=\frac{\text{ mass }}{\text{Density }}$$\frac{1\mathrm{kg}}{2.20\mathrm{lb}}; \frac{2.20\mathrm{lb}}{1\mathrm{kg}}$

Step: 3 Finding volume of iron: (part a)

The volume of iron as

$\text{ volume }=\frac{0.25l\mathrm{b}\times \frac{1\mathrm{kg}}{2.20l\mathrm{g}}\times \frac{1000\mathrm{g}}{1\mathrm{kg}}}{7.86\mathrm{g}/{\mathrm{cm}}^3}$

$Volume=14.46c{m}^3$

Rounding off the value,

$Volume=14c{m}^3.$

Step: 4 Given Information: (Part b)

Given are: $30$ iron nails and $17.6 mL$ of water.

To find the new level of water in milliliters .

Step; 5 New level of water: (Part b)

The volume of nails by 

$\text{ volume }=\frac{30\text{ nails }\times \frac{0.25l\mathrm{b}}{75\mathrm{nails}}\times \frac{1\mathrm{kg}}{2.20l\mathrm{b}}\times \frac{1000\mathrm{g}}{1\mathrm{kg}}}{7.86\mathrm{g}/{\mathrm{cm}}^3\times \frac{1{\mathrm{cm}}^3}{1\mathrm{mL}}}$

$Volume=5.8mL.$

The new level of water is  $= 5.8 mL+17.6 mL$ $=23.4 mL.$

Step: 6 Given Information: (Part c)

Given are: $75$ iron nails and temperature is ${16}^{\circ }C - {125}^{\circ }C$.

To find the heat in joules.

Step: 7 Diffrenece temperature: (Part c)

The heat equation as 

$\text{ Heat }=\text{ mass }\times \mathrm{\Delta }T\times SH$

The Difference temperature as

$\mathrm{\Delta }T=T_{\text{final }}-T_{\text{initial }}$

$\begin{array}{l}\mathrm{\Delta }T={125}^{\circ }\mathrm{C}-{16}^{\circ }\mathrm{C}\\ \mathrm{\Delta }T={109}^{\circ }\mathrm{C}.\end{array}$

Step: 8 Finding heat: (part c)

Substituting in above equation as, 

$\text{ Heat }=0.25l\mathrm{b}\times \frac{1\mathrm{kg}}{2.20l\mathrm{b}}\times \frac{1000\mathrm{g}}{1\mathrm{kg}}\times {109}^{\circ }\mathrm{C}\times \frac{0.452\mathrm{J}}{{\mathrm{g}}^{\circ }\mathrm{C}}$

$Heat=5.598$

Rounding off the values,

$\begin{array}{l}\text{ Heat }=5600\mathrm{J}\\ \mathrm{Hea}t=5.6\times {10}^3\mathrm{J}.\end{array}$

Step: 9 Given Information: (Part d)

Given are: ${25}^{\circ }C$ of melting point and $75$ iron nails and $7.86 g/c{m}^3$ and weight of iron nails is $0.25lb$.

To find the heat in joules.

Step: 10 Weight of nail: (Part d)

The weight of nail as,
$\begin{array}{l}\text{ Weight in }\mathrm{g}\text{ of }1\text{ nail }=1\text{ nails }\times \frac{0.25\mathrm{lb}}{75\text{ nails }}\times \frac{1\mathrm{kg}}{2.20lb}\times \frac{1000\mathrm{g}}{1\mathrm{kg}}\\ \text{ Weight in }\mathrm{g}\text{ of }1\text{ nail }=1.52\mathrm{g}\\ \text{ Weight in }\mathrm{g}\text{ of }1\text{ nail }\approx 1.5\mathrm{g}.\end{array}$

Step: 11 Finding heat: (Part d)

The heat as, 

$\begin{array}{l}\text{ Heat }=1.5\mathrm{g}\times {1510}^{\circ }\mathrm{C}\times \frac{0.452\mathrm{J}}{{\mathrm{g}}^{\circ }\mathrm{C}}\\ \text{ Heat }=1023.78\mathrm{J}\\ \text{ Heat }=1024\mathrm{J}.\end{array}$

The heat fusion as,

$\text{ Heat }=\text{ mass }\times \text{ heat of fusion }$

$\begin{array}{l}\text{ Heat }=1.5\mathrm{g}\times \frac{227\mathrm{J}}{\mathrm{g}}\\ \text{ Heat }=341\mathrm{J}.\end{array}$

Step: 12 Finding total heat: (Part d)

The total heat as

$\begin{array}{l}\text{ Total heat }=1024\mathrm{J}+341\mathrm{J}\\ \mathrm{Total}heat=1365\mathrm{J}\end{array}$

Rounding off the values

$\begin{array}{l}\text{ Total heat }=1.4\mathrm{J}\times {10}^3\mathrm{J}\\ \text{ Total heat }=1400\mathrm{J}.\end{array}$