# 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?

##### 1 Answer

#### Answer:

#### 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

*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

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

#-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")#