# Question 0ce58

Mar 24, 2017

${65.6}^{\circ} \text{C}$

#### Explanation:

The key here is the specific heat of gold.

As you know, a substance's specific heat tells you the amount of heat needed in order to increase the mass of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, you have

${c}_{\text{gold" = "0.126 J g"^(-1)""^@"C}}^{- 1}$

This means that in order to increase the temperature of $\text{1 g}$ of gold by ${1}^{\circ} \text{C}$, you need to supply it with $\text{0.126 J}$ of heat.

Now, John's gold tooth has a mass of $\text{14.7 g}$. You can use the specific heat of gold to calculate the amount of heat needed to increase the temperature of $\text{14.7 g}$ of gold by ${1}^{\circ} \text{C}$

14.7 color(red)(cancel(color(black)("g"))) * "0.126 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "1.8522 J"""^@"C"^(-1)

So, if you supply $\text{1.8522 J}$ of heat to $\text{14.7 g}$ of gold, you will increase its temperature by ${1}^{\circ} \text{C}$. This means that the heat absorbed from the hamburger will cause its temperature to increase by

46.5 color(red)(cancel(color(black)("J"))) * overbrace( (1^@"C")/(1.8522 color(red)(cancel(color(black)("g")))))^(color(blue)("for 14.7 g of gold")) = 25.1^@"C"

You now know that the temperature of the tooth increased by ${25.1}^{\circ} \text{C}$. You can thus say that the final temperature of the tooth will be

T_"final" = 40.51^@"C" + 25.1^@"C" = color(darkgreen)(ul(color(black)(65.6^@"C")))#

The answer is rounded to three sig figs.