Chemistry 3202 Unit 2 Assignment 3 2003-04 page 5
Chemistry 3202 Unit 3 Section 1 Assignment
This is an old assignment you can use for study purposes.
1. Biff and Molly bought a new refrigerator to store water samples for an acid rain project. The water samples, which were all at room temperature, were stored in sealed plastic bottles and then placed in the new fridge.
(2) (a) Assuming that water, the air in the refrigerator chamber, the refrigerator coolant, the heat dissipation coils, and the air in the laboratory are the things that undergo a heat change, identify all the possible universes.
(2) (b) Using the first law of thermodynamics, compare the energy change of the water to the energy change of the air surrounding the heat dissipation coils.
(2) (c) What type of system is the water? Why?
(2) (d) Each water sample goes from 25°C to 5°C in the refrigerator. Classify the energy change for the water samples as endothermic or exothermic. Justify your choice.
(2) (e) Describe the kinetic energy changes that the water molecules undergo in the refrigerator from the time they enter the fridge to the time they reach thermal equilibrium with the air in the fridge.
(2) (f) What is heat and how is it different from temperature?
(2) 2. (a) Platinum is an important catalyst in automobile emissions control systems. A 10.0 g sample of platinum underwent a 5.00°C temperature change resulting in a gain of 6.63 J. Calculate the specific heat capacity of platinum metal.
(2) (b) A freezer pack has a heat capacity of 9.67 kJ/°C. Calculate the heat change of the freezer pack when it is warmed from -18.4°C to 0.0°C.
(5) 3. (a) Complete the “Procedure Items” in the Thought Lab on page 631 of MHR.
(1) (b) Biff and Molly decided to extend the experiment by measuring the temperature changes in two ethanol samples. They collected this data.
Beaker / 1 / 2mass of ethanol / 60.0 g / 60.0 g
Initial Temperature of ethanol / 50.0°C / 37.0°C
Final Temperature of ethanol / 0.0°C / 0.0°C
Mass of Ice added / 30.0 g / 30.0 g
Mass of Ice melted / 21.5 g / 15.92 g
Compare the amounts of ice melted by each ethanol sample. Which sample releases more heat as it cools to 0°C?
(1) (c) Complete the Analysis item in the thought lab.
(2) 4. Sketch and label a constant pressure calorimeter.
5. A lightweight aluminum-silicon-carbon alloy is being tested for use in high performance brakes. The key properties of the alloy that make it suitable for this purpose are rapid dissipation of heat and high melting point.
(4) (a) A 4.00 g sample of the alloy was heated to 1000.00°C – a temperature similar to maximums reached in the high stress braking conditions of F1 car racing. The sample was added to constant pressure calorimeter holding 150.0 mL of water at 24.38°C. The final temperature of the calorimeter water was 28.14°C. Calculate the specific heat capacity of the alloy.
(1) (b) What is the main assumption associated with use of a constant pressure calorimeter in the above problem?
(5) 6. Biff and Molly noticed that beach sand feels very hot on their feet on a sunny summer’s day, so they decided to determine the heat change associated with cooling sand. They used an infrared thermometer gun to measure the sand temperature and got average readings of 54.44°C. Then they added 25.0 g of dry sand to an insulated Tim Horton’s cup containing 100.0 mL of pond water at 18.05°C. Given that sand has a specific heat capacity of 0.703 J/(g°C), and assuming that the specific heat of pond water is 4.184 J/(g°C), calculate the final temperature of the water in the cup.
Chemistry 3202 Unit 3 Section 1 Assignment Key
2003 - 04
1. (a) Universe 1: water = system, air in fridge = surroundings
Universe 2: air in fridge = system; coolant = surroundings
Universe 3: coolant = system; heat dissipation coils = surroundings
Universe 4: heat coils = system; air outside fridge = surroundings.
(b) The first law of thermodynamics states that the heat gained or lost by a system is equal to the heat lost or gained by the surroundings. In this case, the energy lost by the water in the fridge will equal the heat gained by the air outside the fridge.
(c) The water is a closed system because it is bottled. It doesn’t mix with the air in the refrigerator chamber.
(d) The water undergoes an exothermic change. Heat is lost to the surroundings.
(e) The water molecules in the water samples lose kinetic energy. On average, the water molecules are moving (translating and rotating) more slowly at 5°C than at 25°C.
(f) Heat is the form of energy that flows from a warmer to a cooler object when the objects are in thermal contact. Temperature is an indicator of the average kinetic energy of the particles of matter in a sample. Heat is a function of the mass, nature and initial temperature of a substance. Temperature is independent of the mass or nature of a substance.
2. (a) Given: m = 10.0 g, Δt = 5.00°C, q = 6.63 J.
Find: c = ? J/(g°C)
The specific heat capacity of platinum is 0.133 J/g°C.
(b) Given: C = 9.67 kJ/°C. ti = -18.4°C, tf = 0.0°C
Find: q = ? J or kJ
The heat change of the freezer pack is 178 kJ.
3. (a) 1 (a): In part A, the manipulated variable was the mass of the water. In part B, the manipulated variable was the nature of the liquid.
1 (b): The responding variable was the temperature of the substance.
1 (c): The controlled variables were the initial temperatures of the samples tested and the mass of ice added to each sample. Presumably the initial temperatures of the ice were the same too.
2 (a): If ice is added to water samples of different mass, sample with the lower mass will undergo a larger temperature change.
2(b): If ice is added to two different liquids, the liquid with a lower capacity to absorb heat will undergo a larger temperature change.
(b) The warmer ethanol sample loses more heat resulting in a greater mass of ice melted
(c) The three factors that influence the magnitude of a heat change are the mass of a sample, its specific heat capacity, and the temperature change it undergoes.
4.
5. (a) Given: malloy = 4.00 g, ti alloy = 1000.00°C, vwater = 150.0 mL ti water = 24.38°C,
tf = 28.14°C.
Find: calloy = ? J/(g°C)
Strategy:
(1) find heat change for water,
(2) apply first law of thermodynamics,
(3) find specific heat capacity of alloy.
The specific heat capacity of the alloy is 0.607 J/(g°C)
(b) Only the calorimeter water absorbs heat from the alloy. In other words, the assumption is that the other materials in the device do not absorb heat from the sample.
(5) 6. Given: ti sand = 54.44°C, msand = 25.0 g, vwater = 100.0 mL, ti water = 18.05°C.
csand = 0.703 J/(g°C), cwater = 4.184 J/(g°C),
Find: tf = ?°C
Strategy: Set up first law equation and substitute the given data. Rearrange to solve for tf.
The final temperature of the sand and the water is 19.52°C.