Question #b29dc

1 Answer
Feb 2, 2016

Answer:

#DeltaT = 38^@"C"#

Explanation:

The key to understanding this type of problems lies with the substance's specific heat.

For a given substance, its specific heat will tell you how much heat is needed in order to increase the mass of #"1 g"# of that substance by #1^@"C"#.

In your case, the specific heat of gold is said to be

#c = 0.128"J"/("g" ""^@"C")#

This tells you that you need #"0.128 J"# of heat for every gram of the sample and for every #1^@"C""# increase in its temperature!

So, for example, your sample of gold is said to have a mass of #"20 g"#. What if we wanted to increase the temperature of this sample by just #1^@"C"?#

Well, since you need #"0.128 J"# for every gram, and since you only want to increase its temperature by #1^@C"#, it follows that you'd need

#20 xx "0.128 J" = "2.56 J"#

But you ended up using #"96 J"#, which means that the difference between this value and the value given to you went into increasing the sample's temperature!

You can say that

#x xx "2.56 J" = "96 J"#

Here #x# represents the increase in temperature produced by adding that extra heat to the sample. This means that you have

#x = 96/2.56 = 37.5#

The temperature of the coin increased by #37.5^@"C"#.

You actually have an equation that you can use for such problems

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - the amount of heat added / removed
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature

Plugging your values into this equation will once again give you

#DeltaT = q/(m * c)#

#DeltaT = (96 color(red)(cancel(color(black)("J"))))/(20 color(red)(cancel(color(black)("g"))) * 0.128color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 37.5^@"C"#

I'll leave the answer rounded to two sig figs

#DeltaT = color(green)(38^@"C")#