# Question 3d744

Jun 19, 2017

The final temperature of the copper is 18.7 °C.

#### Explanation:

This is a typical calorimetry question.

There are two heat transfers involved:

$\text{heat lost by Cu + heat gained by water} = 0$

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

color(white)(mm)m_1C_1ΔT_1 color(white)(m)+color(white)(mm) m_2C_2ΔT_2 color(white)(mmm)= 0

Let's calculate each heat separately.

q_1 = m_1C_1ΔT_1 = 250. color(red)(cancel(color(black)("g"))) × "0.385 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "100 °C") = "96.25 J·°C"^"-1"(T_text(f) - "100 °C") = 96.25T_text(f) color(white)(l)"J·°C"^"-1" - "9625 J"

q_2 = m_2C_2ΔT_2 = 500 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "15 °C") = "2092 J·°C"^"-1"(T_text(f) - "15 °C") = 2092T_text(f) color(white)(l)"J·°C"^"-1" - "31 380 J"#

Now, we add the two heats and combine like terms.

${q}_{1} + {q}_{2} = 96.25 {T}_{\textrm{f}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J")))·"°C"^"-1" - "9625" color(red)(cancel(color(black)("J"))) + 2092T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - "31 380" color(red)(cancel(color(black)("J}}}} = 0$

$2188 {T}_{\textrm{f}} \textcolor{w h i t e}{l} \text{°C"^"-1" - "41 005} = 0$

${T}_{\textrm{f}} = \text{41 005"/("2188 °C"^"-1") = "18.7 °C}$