If 20 grams of ice at 0°C is added to 120 grams of water at 49°C, what is the temperature of the mixture?

Jun 23, 2018

The final temperature of the mixtire is 31 °C.

Explanation:

There are three heat transfers involved in this calorimetry problem.

$\text{heat to melt ice + heat to warm melt-water + heat lost by hot water} = 0$

$\textcolor{w h i t e}{m m m m} {q}_{1} \textcolor{w h i t e}{m m l l} + \textcolor{w h i t e}{m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m m m m} + \textcolor{w h i t e}{m m m m} {q}_{3} \textcolor{w h i t e}{m m m m l l} = 0$

color(white)(mml)m_1Δ_text(fus)H color(white)(m)+ color(white)(mmm)m_1C_text(s)ΔT_2 color(white)(mmmmm)+ color(white)(mm)m_3C_text(s)ΔT_3color(white)(mmm) = 0

Let's calculate each heat separately.

q_1 = m_1Δ_text(fus)H = 20 color(red)(cancel(color(black)("g"))) × "333.55 J"·color(red)(cancel(color(black)("g"^"-1"))) = "6670 J"

q_2 = m_1C_text(s)ΔT_2 = 20 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "0 °C") = 83.7 T_text(f) color(white)(l)"J·°C"^"-1"

q_3 = m_3C_text(s)ΔT_3 = 120 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g")))·"°C"^"-1" × (T_text(f) - "49 °C") = 502T_text(f) color(white)(l)"J·°C"^"-1" - "24 600 J"

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

${q}_{1} + {q}_{2} + {q}_{3} = 6670 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J" )))+ 83.7T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" + 502T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - "24 600" color(red)(cancel(color(black)("J}}}} = 0$

$586 {T}_{\textrm{f}} \textcolor{w h i t e}{l} \text{°C"^"-1" - "17 900} = 0$

${T}_{\textrm{f}} = \text{17 900"/("586 °C"^"-1") = "31 °C}$