# Question 59f7f

Feb 28, 2017

The equilibrium temperature of the water is 23.0 °C.

#### Explanation:

The guiding principle is the Law of Conservation of Energy:

The sum of all the energy changes must add up to zero.

The formula for the heat $q$ gained or lost by a substance is

color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "

where

$m$ is the mass of the substance.
$c$ is its specific heat capacity.
ΔT = T_"f" - T_"i" is the change in temperature.

In this problem, there are two heat flows.

$\text{Heat lost by copper + Heat gained by water} = 0$

$\textcolor{w h i t e}{m m m m m} {q}_{1} \textcolor{w h i t e}{m m m l l} + \textcolor{w h i t e}{m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m l} = 0$

color(white)(mmm)m_1c_1ΔT_1 color(white)(mml)+color(white)(mmm) m_2c_2ΔT_2color(white)(mmll) = 0

The final temperature of the copper will be the same as the final temperature of the water.

In this problem,

${m}_{1} = \text{5.61 g";color(white)(mmmm) m_2 = "100 g}$

${c}_{1} = \text{0.3844 J·K"^"-1""g"^"-1"; c_2 = "4.184 J·K"^"-1""g"^"-1}$

ΔT_1 = T_"f"color(white)(l) "- 98.8 °C"; color(white)(ll)ΔT_2 = T_"f" color(white)(l)"- 22.6 °C"

m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0

5.61 color(red)(cancel(color(black)("g"))) × 0.3844 color(red)(cancel(color(black)("J·K"^"-1""g"^"-1"))) × (T_"f"color(white)(l) "- 98.8 °C")
+ 100 color(red)(cancel(color(black)("g"))) ×4.184 color(red)(cancel(color(black)("J·K"^"-1""g"^"-1"))) × (T_"f" color(white)(l)"- 22.6 °C") =0#

$2.156 \left({T}_{\text{f"color(white)(l) "- 98.8 °C") + 418.4(T_"f" color(white)(l)"- 22.6 °C}}\right) = 0$

$2.156 {T}_{\text{f" - "213.1 °C" + 418.4T_"f" - "9456 °C}} = 0$

$420.6 {T}_{\text{f" = "9669 °C}}$

${T}_{\text{f" = "9669 °C"/420.6 = "23.0 °C}}$