Question 8ad50

Aug 12, 2016

${28}^{\circ} \text{C}$

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

$\textcolor{b l u e}{- {q}_{\text{copper" = q_"water")" " " "color(orange)("(*)}}}$

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

Your tool of choice here will this equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the 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

${c}_{\text{copper" = "0.386 J g"^(-1)""^@"C}}^{- 1}$

${c}_{\text{water" = "4.186 J g"^(-1)""^@"C}}^{- 1}$

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 ${\text{1 g mL}}^{- 1}$.

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 ${400}^{\circ} \text{C}$ and end up at ${T}_{2}$. On the other hand, the water starts at ${25}^{\circ} \text{C}$ and ends up at ${T}_{2}$.

You can thus say that

$\Delta {T}_{\text{copper" = (T_2 - 400)^@"C}}$

$\Delta {T}_{\text{water" = (T_2 - 25)^@"C}}$

The heat lost by the copper will be equal to

${q}_{\text{copper" = m_"copper" * c_"copper" * DeltaT_"copper}}$

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")))

${q}_{\text{copper" = 9.65 * (T_2 - 400)" J}}$

The heat gained by the water will be equal to

${q}_{\text{water" = m_"water" * c_"water" * DeltaT_"water}}$

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")))#

${q}_{\text{water" = 1046.5 * (T_2 - 25)" J}}$

Now use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to write

$- 9.65 \cdot \left({T}_{2} - 400\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) = 1046.5 * (T_2 - 25)color(red)(cancel(color(black)("J}}}}$

This will get you

$- 9.65 \cdot {T}_{2} + 3860 = 1046.5 \cdot {T}_{2} - 26162.5$

Rearrange to solve for ${T}_{2}$

$1056.15 \cdot {T}_{2} = 30022.5 \implies {T}_{2} = \frac{30022.5}{1056.15} = 28.43$

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{T}_{2} = {28}^{\circ} \text{C}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.