# Question 04da9

Sep 11, 2017

The mass of water originally in the calorimeter was 76.58 g.

#### Explanation:

We must identify the heat transfers that are occurring here.

One is the heat transferred from the hot water as it cools (${q}_{1}$).

Another is the heat transferred to the water in the calorimeter as it warms (${q}_{2}$).

The third is the heat transferred to the calorimeter as it warms (${q}_{3}$).

Per the Law of Conservation of Energy, the sum of the three heat transfers must be zero.

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

The formula for the heat absorbed by or released from a substance is

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

where

$q$ is the quantity of heat
$m$ is the mass of the substance
$c$ is the specific heat capacity of the material
ΔT is the temperature change

This gives us

m_1C_1ΔT_1 + m_2C_2ΔT_2 + C_text(cal)ΔT_3 = 0

We have

${m}_{1} \textcolor{w h i t e}{l l} = \text{49.09 g}$
${C}_{1} \textcolor{w h i t e}{l l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT_1 = T_"f" - T_"i" = "(42.1 - 72.3) °C" = "-30.2 °C"

m_2color(white)(ll) = ?
${C}_{2} \textcolor{w h i t e}{l l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT_2 = T_"f" - T_"i" = "(42.1 - 24.2) °C" = "17.9 °C"

${C}_{\textrm{c a l}} = \text{26.3 J·°C"^"-1}$
ΔT_3 = T_"f" - T_"i" = "(42.1 - 24.2) °C" = "17.9 °C"

q_1 = 49.09 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × ("-30.2" color(red)(cancel(color(black)("°C")))) = "-6203"color(white)(l) color(white)(l)"J"

q_2 = m_2 × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 17.9 color(red)(cancel(color(black)("°C"))) = 74.89m_2color(white)(l) "J·g"^"-1"

${q}_{3} = \text{26.3 J"·color(red)(cancel(color(black)("°C"^"-1"))) × 17.9 color(red)(cancel(color(black)("°C"))) = "470.7 J}$

q_1 + q_2 + q_3 = "-6203" color(red)(cancel(color(black)("J"))) + 74.89m_2color(red)(cancel(color(black)("J")))·"g"^"-1" + 470.7 color(red)(cancel(color(black)("J"))) = 0

$74.89 {m}_{2} \textcolor{w h i t e}{l} \text{g"^"-1" = "6203 - 470.7} = 5732$

m_2 = 5732/("74.89 g"^"-1") = "76.5 g"#