91Ó°ÊÓ

Given,

The density of gold = $19.3\mathrm{g}/{\mathrm{cm}}^3$

Specific heat = $0.129\mathrm{J}/{\mathrm{g}}^{∘}\mathrm{C}$

Melting point =${1064}^{∘}\mathrm{C}$

The weight of the nugget is$20.17lb$

It has four significant figures in the measurement of the weight of the nugget because zero and non-zero digits present between non zero digits are significant.

The weight of the nugget is $20.17lb$

now, the weight in kg is,

We know,

The density of gold is$19.3\mathrm{g}/{\mathrm{cm}}^3$

and mass is role="math" localid="1652073603365" $9.17kg=9170g$

Volume = $\frac{mass}{density}$

$⇒\frac{1970}{19.3}=475c{m}^3$

We know,

the melting point of gold = ${1064}^{∘}\mathrm{C}$

the melting point of gold in Fahrenheit is$T_{\mathrm{F}}=1.8\left(T_{\mathrm{C}}\right)+32$

$=1.8\left(1064\right)+32={1947}^{o}F$

the melting point of gold in kelvin is $T\left(\mathrm{K}\right)=T\left({}^{∘}\mathrm{C}\right)+273$

$=1064+273={1337}^{o}K$

We know,

Mass = $9.17kg$

specific heat $c=0.129\mathrm{J}/{\mathrm{g}}^{∘}\mathrm{C}$

Change in temperature$∆T={500}^{∘}\mathrm{C}\text{to}{1064}^{∘}\mathrm{C}={564}^{\mathrm{o}}\mathrm{C}$

$1\mathrm{cal}=4.184\mathrm{J}$

The calories required are,

$Q=mc\mathrm{Δ}T$

$\frac{⇒9170×564×0.129}{4.184×1000}=159kcal$

We know,

price of gold = $\mathrm{\$}42.06\text{per gram}$

Price of nugget

$$\begin{gathered} =9170×42.06 \\ =\$386000 \end{gathered}$$