Question

Question #0ce58

Answer 1

`65.6^@"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 `"1 g"` of that substance by `1^@"C"`.

In your case, you have

`c_"gold" = "0.126 J g"^(-1)""^@"C"^(-1)`

This means that in order to increase the temperature of `"1 g"` of gold by `1^@"C"`, you need to supply it with `"0.126 J"` of heat.

Now, John's gold tooth has a mass of `"14.7 g"`. You can use the specific heat of gold to calculate the amount of heat needed to increase the temperature of `"14.7 g"` of gold by `1^@"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 `"1.8522 J"` of heat to `"14.7 g"` of gold, you will increase its temperature by `1^@"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^@"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.