Question

In one box of nails, there are $75$iron nails weighing $0.250lb$. The density of iron is $7.86g/c{m}^3$. The specific heat of iron is $0.425J/g^{\circ }C$. 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 $17.6mL$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?

Short 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 $V=23.4mL.$

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-by-step solution

Step: 1 Given information: (Part a) 

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

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$17.6mL$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 $=5.8mL+17.6mL$$=23.4mL.$

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 $7.86g/c{m}^3$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

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