Question

A 1.50-kg iron horseshoe initially at 600.0°C is dropped into a bucket containing 20.0 kg of water at 25.0°C. What is the final temperature?

Answer 1

`30.1^@"C"`

Explanation:

The idea here is that the heat lost by the metal will be equal to the heat gained by the water.

In order to be able to calculate the final temperature of the iron + water system, you need to know the [specific

heats](http://socratic.org/chemistry/thermochemistry/specific-heat) of water and of iron, respectively, which are

listed as being equal to

`c_"water" = 4.18"J"/("g" ""^@"C")" "` and `" "c_"iron" = 0.45"J"/("g" ""^@"C")`

The equation that establishes a relationship between heat lost/gained and change in temperature looks like this

`color(blue)(q = m * c * DeltaT)" "`, where

`q` - heat lost or gained
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as final temperature minus initial temperature

In your case, you know that

`color(blue)(-q_"iron" = q_"water")`

The minus sign is used here because heat lost carries a negative sign. Let's say that the final temperature

of the iron + water system will be `T_"f"`.

You can say that the changes in temperature for the iron and for the water will be

`DeltaT_"iron" = T_"f" - 600.0^@"C"" "` and `" "DeltaT_"water" = T_"f" - 25.0^@"C"`

This means that you will have

`-m_"iron" * c_"iron" * DeltaT_"iron" = m_"water" * c_"water" * DeltaT_"water"`

which is equivalent to

`-m_"iron" * c_"iron" * (T_"f" - 600.0^@"C") = m_"water" * c_"water" * (T_"f" - 25.0^@"C")`

Notice that the specific heats for these two substances is given per gram. This means that you will have to

convert the two masses from kilograms to grams

`"For Fe: " 1.50color(red)(cancel(color(black)("kg"))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = "1500 g"`

`"For H"_2"O: " 20.0color(red)(cancel(color(black)("kg"))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = "20000 g"`

Plug your values into the equation and solve for `T_"f"`

`-1500color(red)(cancel(color(black)("g"))) * 0.45color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g")))color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 600.0)color(red)(cancel(color(black)(""^@"C"))) = 20000color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25.0)color(red)(cancel(color(black)(""^@"C")))`

`405000 - 675 * T_"f" = -2090000 + 83600 * T_"f"`

`82925 * T_"f" = 2495000 implies T_"f" = 2495000/82925 = 30.09^@"C"`

Rounded to three sig figs, the

answer will be

`T_"f" = color(green)(30.1^@"C")`