# Question #8ad50

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the heat **lost** by the piece of copper will be * equal* to the heat

**absorbed**by the water, as stated by the hint given to you in the problem

#color(blue)(-q_"copper" = q_"water")" " " "color(orange)("(*)")#

The *minus sign* is used here because **heat lost** carries a negative sign.

Your tool of choice here will this equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

The **specific heat** of copper and the **specific heat** of water are listed as

#c_"copper" = "0.386 J g"^(-1)""^@"C"^(-1)#

#c_"water" = "4.186 J g"^(-1)""^@"C"^(-1)#

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html

Now, the problem doesn't provide you with a *mass* of water; instead, you get a *volume* of water. When the **density** of the water is not given to you, you can approximate it to be equal to

This means that your volume of water would have a mass of

#250 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "250 g"#

So, the piece of copper will start at

You can thus say that

#DeltaT_"copper" = (T_2 - 400)^@"C"#

#DeltaT_"water" = (T_2 - 25)^@"C"#

The heat **lost** by the copper will be equal to

#q_"copper" = m_"copper" * c_"copper" * DeltaT_"copper"#

#q_"copper" = 25 color(red)(cancel(color(black)("g"))) * "0.386 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_2 - 400)color(red)(cancel(color(black)(""^@"C")))#

#q_"copper" = 9.65 * (T_2 - 400)" J"#

The heat **gained** by the water will be equal to

#q_"water" = m_"water" * c_"water" * DeltaT_"water"#

#q_"water" = 250 color(red)(cancel(color(black)("g"))) * "4.186 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_2 - 25)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = 1046.5 * (T_2 - 25)" J"#

Now use equation

#-9.65 * (T_2 - 400) color(red)(cancel(color(black)("J"))) = 1046.5 * (T_2 - 25)color(red)(cancel(color(black)("J")))#

This will get you

#-9.65 * T_2 + 3860 = 1046.5 * T_2 - 26162.5#

Rearrange to solve for

#1056.15 * T_2 = 30022.5 implies T_2 = 30022.5/1056.15 = 28.43#

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

#color(green)(|bar(ul(color(white)(a/a)color(black)(T_2 = 28^@"C")color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but you should leave it rounded to one sig fig, the number of sig figs you have for the initial temperature of the copper.