# What is the heat capacity C_(cal) of the dry calorimeter device?

## A piece of iron (specific heat capacity = 0.452 J·g–1·˚C–1) with a mass of 54.3218 g is heated in a water bath until it reaches a temperature of 98.2 ˚C. The iron is then transferred into a calorimeter (nested Styrofoam cups) that contains 68.5314 g of water initially at 22.1 ˚C. The final temperature of the entire system is 27.5 ˚C. The specific heat capacity of water is 4.184 J·g–1·˚C–1

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Jan 8, 2018

Warning! Long Answer. ${C}_{\textrm{C a l}} = \text{35 J·°C"^"-1}$

#### Explanation:

We must identify the heat transfers that are occurring here.

One is the heat lost by the iron as it cools down (${q}_{1}$).

A second is the heat gained by the water as it warms up (${q}_{2}$).

A third is the heat gained by the calorimeter as it warms up (${q}_{3}$).

Thus, the heat lost by the iron goes into warming up the water and the calorimeter.

One formula for the heat $q$ absorbed by or released from 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 the specific heat capacity of the material
ΔT is the temperature change

A useful formula for the calorimeter is

color(blue)(bar(ul(|color(white)(a/a)q = C_text(Cal)ΔTcolor(white)(a/a)|)))" "

where

"C_text(Cal) is the "calorimeter constant" (the heat capacity of the calorimeter)

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

$\textcolor{w h i t e}{m m l} {q}_{1} \textcolor{w h i t e}{m l} + \textcolor{w h i t e}{m m l} {q}_{2} \textcolor{w h i t e}{m l l} + \textcolor{w h i t e}{m m l} {q}_{3} \textcolor{w h i t e}{m l l} = 0$
m_1C_1ΔT_1 + m_2C_2ΔT_2 + C_text(Cal)ΔT_text(Cal) = 0

In this problem

${m}_{1} \textcolor{w h i t e}{l l} = \text{54.3218 g}$
${C}_{1} \textcolor{w h i t e}{l l} = \text{0.452 J·°C"^"-1""g"^"-1}$
ΔT_1 = T_"f" - T_"i" = "(27.5 - 98.2) °C" = "-70.7 °C"

${m}_{2} \textcolor{w h i t e}{l l} = \text{68.5314 g}$
${C}_{2} \textcolor{w h i t e}{l l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT_2 = T_"f" - T_"i" = "(27.5 - 22.1) °C" = "-5.4 °C"

"C_text(Cal) = ?
ΔT_3 = T_"f" - T_"i" = "(27.5 - 22.1) °C" = "-5.4 °C"

q_1 = 54.3218color(red)(cancel(color(black)("g"))) × "0.452 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × ("-70.7" color(red)(cancel(color(black)("°C")))) = "-1736 J"

q_2 = 68.5314 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 5.4color(red)(cancel(color(black)("°C"))) = "1550 J"

q_3 = C_text(Cal) × 5.4 °C = 5.4C_text(Cal)color(white)(l)"°C"

${q}_{1} + {q}_{2} + {q}_{3} = \text{-1736 J +1550 J + 5.4"C_text(Cal)color(white)(l)"°C} = 0$

$5.4 {C}_{\textrm{C a l}} \textcolor{w h i t e}{l} \text{°C = 190 J}$

${C}_{\textrm{C a l}} = \text{190 J"/"5.4 °C" = "35 J·°C"^"-1}$

Note: The answer can have only two significant figures because that is all we have for ΔT_2 and ΔT_3.

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