Question

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)?

Answer 1

`"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)|)))`