# 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)?

Apr 11, 2016

$\text{680 kJ}$

#### Explanation:

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

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of $\text{1 g}$ of this substance by ${1}^{\circ} \text{C}$.

This is of course equivalent to saying that specific heat tells you how much heat is released when the temperature of $\text{1 g}$ of a substance decreases by ${1}^{\circ} \text{C}$.

In your case, iron's specific heat tells you that decreasing the temperature of a $\text{1-g}$ sample of iron by ${1}^{\circ} \text{C}$ will give off $\text{0.46 J}$ of heat.

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

The iron starts at ${1500}^{\circ} \text{C}$ and ends up at ${25}^{\circ} \text{C}$, which means that its temperature decreases by

$| \Delta T | = | {25}^{\circ} \text{C" - 1500^@"C"| = 1475^@"C}$

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 $\text{0.46 J}$ for every gram and for every ${1}^{\circ} \text{C}$ decrease in temperature. This means that you can break up the calculations into two parts

• calculate the heat given off when $\text{1 g}$ of iron undergoes a ${1475}^{\circ} \text{C}$ 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

$\text{678.5 J"color(red)(cancel(color(black)("g"^(-1)))) xx 10^3 color(red)(cancel(color(black)("g"^(-1)))) = "678,500 J}$

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

• calculate the heat given off when ${10}^{3} \text{g}$ of iron undergo a ${1}^{\circ} \text{C}$ 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

$\text{460 J" color(red)(cancel(color(black)(""^@"C"^(-1)))) * 1475 color(red)(cancel(color(black)(""^@"C"))) = "678,500 J}$

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - 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

$q = {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 0.46"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25 - 1500)color(red)(cancel(color(black)(""^@"C}}}}$

$q = - \text{678,500 J}$

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)|)))