Molten iron is extremely hot, averaging about 1,500°C The specific heat of iron is 0.46 J/g°C. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25°C)?

1 Answer
Apr 11, 2016

Answer:

#"680 kJ"#

Explanation:

The key to this problem lies with the specific heat of iron, which is said to be equal to #"0.46 J g"^(-1)""^@"C"^(-1)#.

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

This is of course equivalent to saying that specific heat tells you how much heat is released when the temperature of #"1 g"# of a substance decreases by #1^@"C"#.

In your case, iron's specific heat tells you that decreasing the temperature of a #"1-g"# sample of iron by #1^@"C"# will give off #"0.46 J"# of heat.

The first thing to do here is convert the mass of the sample from kilograms to grams by using the conversion factor

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#

In your case, you have

#1 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 10^3"g"#

The iron starts at #1500^@"C"# and ends up at #25^@"C"#, which means that its temperature decreases by

#|DeltaT| = |25^@"C" - 1500^@"C"| = 1475^@"C"#

#color(red)(!!!)color(white)(a)#SIDE NOTE I've used the absolute value of the change in temperature because heat given off carries a negative sign, but it's reported with a positive sign.

So, iron gives of #"0.46 J"# for every gram and for every #1^@"C"# decrease in temperature. This means that you can break up the calculations into two parts

  • calculate the heat given off when #"1 g"# of iron undergoes a #1475^@"C"# decrease in temperature

#1475color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.46 J g"^(-1)/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("heat given of per gram")) = "678.5 J"#

  • multiply this value by the total mass of the sample

#"678.5 J"color(red)(cancel(color(black)("g"^(-1)))) xx 10^3 color(red)(cancel(color(black)("g"^(-1)))) = "678,500 J"#

Notice that you can get the same result by using an alternative route

  • calculate the heat given off when #10^3"g"# of iron undergo a #1^@"C"# decrease in temperature

#10^3color(red)(cancel(color(black)("g"))) * overbrace(("0.46 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat given off per 1"^@"C")) = "460 J"^@"C"^(-1)#

  • multiply this by the total decrease in temperature

#"460 J" color(red)(cancel(color(black)(""^@"C"^(-1)))) * 1475 color(red)(cancel(color(black)(""^@"C"))) = "678,500 J"#

Expressed in kilojoules and rounded to two sig figs, the answer will be

#"heat given off" = color(green)(|bar(ul(color(white)(a/a)"680 kJ"color(white)(a/a)|)))#

ALTERNATIVE APPROACH

You can get the same result by using the equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#, where

#q# - the amount of heat gained / lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

This equation will produce a negative value for #q#. Keep in mind that the negative sign is used to symbolize heat lost, which is why you'd still report the result with a positive sign

Plug in your values to get

#q = 10^3 color(red)(cancel(color(black)("g"))) * 0.46"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25 - 1500)color(red)(cancel(color(black)(""^@"C")))#

#q = -"678,500 J"#

This tells you that heat is being given off, hence the minus sign. To report how much heat is given off, drop the minus sign to get

#"heat given off" = color(green)(|bar(ul(color(white)(a/a)"680 kJ"color(white)(a/a)|)))#