Name _______________________
Date _______
Period 1 2 3 4 5 6
Regular Chemistry Spring Performance Task
Practice KEY
A calorimetry experiment is carried out to determine the heat of solution for potassium
chloride. In a typical experiment, 100. mL of water at 25.0 °C is placed in the polystyrene cup
and the initial temperature of the water recorded. 5.05 g potassium chloride, KCl, is added to
the water while stirring. The temperature of the solution falls and the minimum temperature
achieved is recorded as 21.7 °C
Data Table
VH2O
MKCl
Tinitial
Tfinal
Cwater
=
=
=
=
=
volume of water
mass of KCl(s)
initial temperature
final temperature
Specific heat water
=
=
=
=
=
100. mL
5.05 g
25.0oC
21.7oC
4.184 J/g oC
1. Fill in the data table above
2. Write a balanced equation for the dissolution of
potassium chloride
KCl(s)  K+(aq) + Cl-(aq)
3. Draw a picture or use a paragraph to describe the interaction between the water
molecules and ions after being fully dissolved.
4. What is the mass of the solution?
1𝑔
100. 𝑚𝐿 𝑥 1 𝑚𝐿 = 100. 𝑔 𝐻2 𝑂
The mass of the solute is considered insignificant and ignored
5. Calculate T for the solution
21.7 oC -25.0 oC = -3.3 oC
6. Assuming the polystyrene foam cup is well insulated and the specific heat capacity of water
is 4.18 J/goC , determine the heat of solution.
q = m x c x ΔT
𝑞 = 100 𝑔 (4.18
𝐽
𝑔𝐶
) (21.7𝐶 − 25.0𝐶)
q = -1380 J
7. Is this reaction endothermic or exothermic? How do you know this? Reference data from
lab to back up your claim.
In the experiment potassium chloride was added to water. After completely dissolving, the
temperature of the solution fell from 25𝐶 to 21.7𝐶. A negative change in temperature
demonstrates that heat was absorbed. Because the reaction is endothermic, ΔHsoln must be
positive (+).
http://www.tutorvista.com/content/physics/physics-i/heat/calorimetry.php
8. A student places two 10.0 g samples of metal, at 25oC each, are exposed to 20 J of heat.
One sample is Lead (CPb = 0.129 J/gC) and the other is silver (CAg= 0.235 J/gC).
a. Which one will have a higher final temperature? _Lead_____________________
b. Explain your choice *hint* no math is required for this one:
“Specific heat” refers to the needed heat to adjust the temperature of a single unit of a
substance’s mass by one degree. The higher the specific heat of a substance, the more energy it
takes to change the temperature. We say that substances with high specific heat are resistant
to temperature change. Substances with low specific heat are sensitive to temperature change.
Consider the water and sand at the beach. The water has a higher heat capacity (4.18 J/g oC)
than the sand (0.83 J/goC) so it feels cool to swim in whereas, the sand is hot to the touch.
9. Two samples of copper (50 gram and 25 gram) are exposed to 10 J of heat energy. Both
started at 10oC. The specific heat of copper is 0.385 J/gC.
a. Which will have a higher final temperature? _The 25 gram sample______________
b. Explain your choice *hint* no math is required for this one:
Heat capacity (C) does change with mass. However, specific heat is the heat capacity per
unit mass (c=Cm). If you double the amount of mass in your system, you've doubled its heat
capacity, but you've kept the specific heat the same.
10. A calorimetry experiment is carried out to determine th identify of an unknown metal. A
25.6 g piece of metal was taken from a beaker of boiling water at 100.0 °C and placed directly
into a calorimeter holding 100.0 mL of water at 25.0 °C. Given that the final temperature at
thermal equilibrium is 26.2 °C, determine the specific heat capacity of the metal.
1) Determine the mass of the water
100.0 𝑚𝐿 𝑥
1𝑔
= 100.0 𝑔
1 𝑚𝐿
2) Determine the energy lost by the water
qwater = m cwater T
qwater= = (100.0g) (4.184 J/gC) (26.2 – 25.0) C
qwater = +502.08 J
extra sig fig!
3) Determine the energy gained by the metal
qlost = qgained
qmetal = -502.08 J
4) Determine the heat capacity of the metal
qmetal = mmetal cmetal T
-502.08 J = 25.6 g (cmetal) (26.2-100) C
-502.08 J = 25.6 g (cmetal) (-73.8) C
-502.08 J = -1889.3 gC (cmetal)
-502.08 J = cmetal
-1889.3 gC
Cmetal = 0.266 J/ gC
4) Identify the metal using the list of specific heats found at the link below:
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
Niobium is closest at 0.27 J/ gC
http://www.chemteam.info/Thermochem/Determine-Specific-Heat.html
5) Describe how the Law of Conservation of Energy applies to calorimetry. This explanation
should explain how we were able to get our answer for #3 of this problem.
Calorimetry is the measurement of the amount of heat evolved or absorbed in a chemical
reaction, change of state, or formation of a solution.
We know that heat flows from a hot substance to a cold substance. In Calorimetry, a hot
substance and a cold substance are mixed together. The hot substance loses heat to the cold
substance until their temperatures are equalized.
According to law of conservation of energy, heat lost by the hot substance = heat gained by the
cold substance. This is the principle of calorimetry.
http://www.tutorvista.com/content/physics/physics-i/heat/calorimetry.php