# Question #8b482

##### 1 Answer

#### Explanation:

The idea here is that the heat **given off** by the piece of iron as it cools will be **equal** to the heat **absorbed** by the water as it warms.

#color(blue)(ul(color(black)(-q_"iron" = q_"water")))" " " "color(darkorange)("(*)")#

Keep in mind that we use a minus sign here because, by convention, heatgiven offcarries a negative sign

Before moving forward, look up the **specific heat** values of water and iron, respectively. You'll find them listed as

#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#

#c_"iron" = "0.444 J g"^(-1)""^@"C"^(-1)#

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

Now, your tool of choice here will be the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

#q# is the heat lost or gained by the substance#m# is themassof the sample#c# is thespecific heatof the substance#DeltaT# is thechange in temperature, defined as the difference between thefinal temperatureand theinitial temperatureof the sample

If you take **final temperature** of the iron + water system, you can say that the change in temperature for the piece of iron will be

#DeltaT_"iron" = T_f - 95^@"C"#

Similarly, the change in temperature for the water will be

#DeltaT_"water" = T_f - 22^@"C"#

Convert the two masses from *kilograms* to *grams*

#"2.4 kg" = 2.4 * 10^(3) color(white)(.)"g" = "2400 g"#

#"0.6 kg" = 0.6 * 10^3 color(white)(.)"g" = "600 g"#

You can thus say that the heat given off by the piece of iron will be equal to

#q_"iron" = 600 color(red)(cancel(color(black)("g"))) * "0.444 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 95)color(red)(cancel(color(black)(""^@"C")))#

#q_"iron" = 266.4 * (T_f - 95)#

The heat absorbed by the water will be equal to

#q_"water" = 2400 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 22)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = 10041.6 * (T_f - 22)#

You can now use equation

#-266.4 * (T_f - 95) = 10041.6 * (T_f - 22)#

This will be equivalent to

#-266.4T_f + 25308 = 10041.6T_f - 220915.2#

Rearrange to get

#(10041.6T_f + 225.4) * T_f = 220915.2 + 25308#

You will thus have

#T_f = (220915.2 + 25308)/(10041.6 + 226.4) = 23.98#

You can thus say that the final temperature of the iron + water system will be

#color(darkgreen)(ul(color(black)(T_f = 24^@"C")))#

The answer is rounded to two **sig figs**.