If a sample of chloroform is initially at 25°C, what is its final temperature of 150.0 g of chloroform absorbs 1.0 kilojoules of heat, and the specific heat of chloroform is 0.96 J/g°C?

1 Answer
Jan 23, 2016

Answer:

#32^@"C"#

Explanation:

!! SHORT ANSWER !!

This problem is a pretty straightforward application of the equation

#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, calculated as the difference between the final temperature and the initial temperature

All you'd have to do here is rearrange this equation to solve for #DeltaT# and plug in your values - do not forget to convert the heat from kilojoules to joules

#DeltaT = q/(m * c)#

#DeltaT = (1.0 * 10^3 color(red)(cancel(color(black)("J"))))/(150.0 color(red)(cancel(color(black)("g"))) * 0.96color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) * ""^@"C")) = 6.94^@"C"#

The final temperature of the sample will thus be

#T_"final" = T_"initial" + DeltaT#

#T_"final" = 25^@"C" + 6.94^@"C"#

#T_"final" = color(green)(32^@"C") -># rounded to two sig figs

!! AN ALTERNATIVE APPROACH !!

I want to show you how to solve such questions without using that equation.

As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In your case, chloroform is said to have a specific heat of

#c = 0.96"J"/("g" ""^@"C")#

So, what does that tell you?

In order to increase the temperature of #"1 g"# of chloroform by #1^@"C"#, you need to provide it with #"0.96 J"# of heat.

Now look at the data given to you. Focus on the mass of the sample first.

Now, if you need to provide #"0.96 J"# of heat in order to increase the temperature of #"1 g"# of chloroform by #1^@"C"#, how much heat will be needed in order to increase the temperature of #"150.0 g"# of chloroform by #1^@"C"#?

Since you need #"0.96 J"# per gram, you will need

#150.0 xx "0.96 J" = "144 J"#

Notice that the problem tells you that you add #"1 kJ"# of heat to the sample. This means that the ratio between the given heat and the amount of heat needed to increase the temperature of the sample by #1^@"C"# will represent the actual increase in temperature.

You can thus say that

#1.0 color(red)(cancel(color(black)("kJ"))) * (10^3 color(red)(cancel(color(black)("J"))))/(1 color(red)(cancel(color(black)("kJ")))) * (1^@"C")/(144 color(red)(cancel(color(black)("J")))) = 6.94^@"C"#

The final temperature of the sample will thus be

#T_"final" = 25^@"C" + 6.94^@"C" = 31.94^@"C"#

Rounded to two sig figs, the answer will once again be

#T_"final" = color(green)(32^@"C")#