Question #0ce58

1 Answer
Mar 24, 2017

Answer:

#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.