# 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

#### Answer:

#### Explanation:

The key to this problem lies with the **specific heat** of iron, which is said to be equal to

As you know, a substance's **specific heat** tells you how much heat is needed to increase the temperature of

This is of course equivalent to saying that specific heat tells you how much heat is released when the temperature of

In your case, iron's specific heat tells you that decreasing the temperature of a

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

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

**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 **for every** gram and **for every**

*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

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

This equation will produce a **negative value** for **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)|)))#