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

A piece of iron (specific heat capacity = #"0.452 J"cdot"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"cdot"g"^(-1)""^@ "C"^(-1)#

1 Answer
Jan 7, 2018

Answer:

Warning! Long Answer. #C_text(Cal) = "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.

#color(white)(mml)q_1 color(white)(ml)+color(white)(mml)q_2 color(white)(mll)+ color(white)(mml)q_3 color(white)(mll)= 0#
#m_1C_1ΔT_1 + m_2C_2ΔT_2 + C_text(Cal)ΔT_text(Cal) = 0 #

In this problem

#m_1color(white)(ll) = "54.3218 g"#
#C_1 color(white)(ll)= "0.452 J·°C"^"-1""g"^"-1"#
#ΔT_1 = T_"f" - T_"i" = "(27.5 - 98.2) °C" = "-70.7 °C"#

#m_2color(white)(ll) = "68.5314 g"#
#C_2 color(white)(ll)= "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 = "-1736 J +1550 J + 5.4"C_text(Cal)color(white)(l)"°C"= 0#

#5.4C_text(Cal)color(white)(l)"°C = 190 J"#

#C_text(Cal) = "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#.