# What is the final temperature if you place 501 g of copper at 89.5 °C in a calorimeter containing 159 g of water at 22.8 °C?

## $\text{Specific heat capacity of copper" = "0.387 J·g"^"-1""°C"^"-1}$ $\text{Specific heat capacity of water" = "4.184 J·g"^"-1""°C"^"-1}$ $\text{Heat capacity of calorimeter" = "10.0 J/°C}$

Jun 4, 2017

The final temperature of the system is 37.7 °C.

#### Explanation:

This is a typical calorimetry question.

There are three heats involved:

$\text{heat lost by Cu + heat gained by water"color(white)(l) "+ heat gained by calorimeter} = 0$

${q}_{1} + {q}_{2} + {q}_{3} = 0$

m_1C_1ΔT_1 + m_2C_2ΔT_2 + C_3ΔT_3 = 0

Let's calculate each heat separately.

q_1 = m_1C_1ΔT_1 = 501 color(red)(cancel(color(black)("g"))) × "0.387 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "89.5 °C") = "193.9 J·°C"^"-1"(T_text(f) - "89.5 °C") = 193.9T_text(f) color(white)(l)"J·°C"^"-1" - "17 353 J"

q_2 = m_2C_2ΔT_2 = 159 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "22.8 °C") = "665.3 J·°C"^"-1"(T_text(f) - "22.8 °C") = 665.3T_text(f) color(white)(l)"J·°C"^"-1" - "15 168 J"

q_3 = C_3ΔT_3 = "10.0 J"·"°C"^"-1" × (T_text(f) - "22.8 °C") = 10.0T_text(f) color(white)(l)"J·°C"^"-1" - "228 J"

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

${q}_{1} + {q}_{2} + {q}_{3} = 193.9 {T}_{\textrm{f}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J")))·"°C"^"-1" - "17 353" color(red)(cancel(color(black)("J"))) + 665.3T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - "15 168" color(red)(cancel(color(black)("J"))) +10.0T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - 228 color(red)(cancel(color(black)("J}}}} = 0$

$869.2 {T}_{\textrm{f}} \textcolor{w h i t e}{l} \text{°C"^"-1" - "32 749} = 0$

${T}_{\textrm{f}} = \text{32 749"/("869.2 °C"^"-1") = "37.7 °C}$