Question

How much heat is given off when a 24.7 kg iron ingot is cooled from 880°C to 13°C?

The specific heat of iron is 0.107 cal/g * °C.

Answer 1

`2.3* 10^6"cal"`

Explanation:

Notice that the problem provides you with the mass of the iron ingot, but that it's being expressed in kilograms.

Since the specific heat of iron is given in joules per gram Celsius, `"J g"^(-1) ""^@"C"^(-1)`, you must convert the mass of the sample from kilograms to grams before doing anything else

`24.7 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 24.7 * 10^3"g"`

Now, you can use the specific heat of iron to figure out how much heat would be released when the temperature of `24.7 * 10^3"g"` of iron decreases by `1^@"C"`

`24.7 * 10^3 color(red)(cancel(color(black)("g"))) * overbrace("0.107 cal"/(1color(red)(cancel(color(black)("g"))) ""^@"C"))^(color(purple)("the specific heat of iron")) = 2.643 * 10^3"cal" ""^@"C"^(-1)`

This tells you that in order to cause a `1^@"C"` decrease in temperature to a `24.7 * 10^3"g"` sample of iron, you must remove `2.643 * 10^3"cal"` of heat.

In your case, the temperature of the ingot must decrease by

`"change in temperature" = |13^@"C" - 880^@"C"| = 867^@"C"`

You can thus say that when the temperature of `2.643 * 10^3"g"` of iron decreases by `867^@"C"`, the process releases

`867 color(red)(cancel(color(black)(""^@"C"))) * overbrace((2.643 * 10^3"cal")/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("for 2.643" * 10^3"g of iron")) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.3 * 10^6"cal")color(white)(a/a)|)))`

The answer is rounded to two sig figs.