Water has an unusually high specific heat capacity Water has an unusually high specific heat capacity. Which one of the following statements logically follows from this fact? Compared to other substances, hot water causes severe burns because it is a good conductor of heat. Compared to other substances, water will quickly warm up to high temperatures when heated. Compared to other substances, it takes a considerable amount of heat for a sample of water to change its temperature by a small amount.

Answer: C A substance with a high specific heat capacity is a substance that requires a relative large quantity of heat to cause a small temperature change. Because of this, water does not change its temperature as rapidly as other substances that are heated in the same manner; choice B does not logically follow. Specific heat capacity should not be confused with thermal conductivity. Thermal conductivity is the measure of the ability of a substance to conduct heat; choice A has to do with thermal conductivity.

2. Explain why large bodies of water such as Lake Michigan can be quite chilly in early July despite the outdoor air temperatures being near or above 90°F (32°C).

Lake Michigan is a body of water with a large m value and a large C value. It would take a lot of solar energy absorption to increase its temperature from the cold wintry temperatures to the higher summertime temperatures. It may take a couple of months of summer before the heating of the large mass of water is "complete."

3. The table below describes a thermal process for a variety of objects (indicated by red, bold-faced text). For each description, indicate if heat is gained or lost by the object, whether the process is endothermic or exothermic, and whether Q for the indicated object is a positive or negative value.

4. An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styrofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc.

The water warms up and the energy it gains is equal to the energy lost by the metal. The quantity of energy gained by the water can be calculated as Qwater = m•Cwater•ΔT = (50.0 g)•(4.18 J/g/°C)•(28.1°C-27.0°C) = 229.9 J Now this 229.9 J is equal to the -Qmetal. The specific heat capacity of the metal can be calculated by setting -229.9 J equal to m•C•ΔT. C = -229.9 J/(11.98 g)/(28.1°C - 78.4°C) = 0.382 J/g/°C

5. Jake grabs a can of soda from the closet and pours it over ice in a cup. Determine the amount of heat lost by the room temperature soda as it melts 61.9 g of ice (ΔHfusion = 333 J/g).

Answer: 20. 6 kJ Use the equation Q = m•ΔHfusion where m=61 Answer: 20.6 kJ Use the equation Q = m•ΔHfusion where m=61.9 g and ΔHfusion=333 J/g. Conversion to kiloJoule is of course optional.

6. The heat of sublimation (ΔHsublimation) of dry ice (solid carbon dioxide) is 570 J/g. Determine the amount of heat required to turn a 5.0-pound bag of dry ice into gaseous carbon dioxide. (Given: 1.00 kg = 2.20 lb)

Answer: 1300 kJ (rounded from 1295 kJ) mdry ice = 5. 0 lb•(1. 00 kg/2 Answer: 1300 kJ (rounded from 1295 kJ) mdry ice = 5.0 lb•(1.00 kg/2.2 lb) = 2.2727 kg Now that the mass of dry ice is known, the Q value can be determined. Again, attention must be given to units. Since the mass is known in kilogram, it would be useful to express the heat of sublimation in kJ/kg. So 570 J/g is equivalent to 570 kJ/kg. And so the answer is calculated as Q = mdry ice • ΔHsublimation-dry ice Q = (2.2727 kg)•(570 kJ/kg) = 1295 kJ Q = ~1300 kJ (rounded to two significant digits)

7. Determine the amount of heat required to increase the temperature of a 3.82-gram sample of solid para-dichlorobenzene from 24°C to its liquid state at 75°C. Para-dichlorobenzene has a melting point of 54°C, a heat of fusion of 124 J/g and specific heat capacities of 1.01 J/g/°C (solid state) and 1.19 J/g/°C (liquid state).

Answer: 680 J (rounded from 684 Answer: 680 J (rounded from 684.9 J) This problem requires three steps - calculating the Q1 for raising the temperature of para-dichlorobenzene (abbreviated as PDCB for the remainder of the problem) to 54°C (the melting point), calculating the Q2 for melting the PDCB, and calculating the Q3 for raising the temperature of the liquid PDCB to 75°C. Q1 =(3.82 g)•(1.01 J/g/°C)•(54°C-24°C) = 115.7 J Q2 =(3.82 g)•(124 J/g) = 473.7 J Q3 =(3.82 g)•(1.19 J/g/°C)•(75°C-54°C) = 95.5 J Qtotal = Q1 + Q2 + Q3 = 684.9 J