**January 2023**

The particles in which sample have the *lowest *average kinetic energy?

(1) 50. g of sulfur at 273 K

(2) 40. g of aluminum at 298 K

(3) 30. g of sulfur at 303 K

(4) 20. g of aluminum at 323 K

Solution: Temperature is the measure of the average kinetic energy. Since choice 1 has the lowest temperature, it should have the lowest average kinetic energy as well.

Answer:1

**June 2022**

38 What is the amount of heat required to melt 43 grams of solid magnesium oxide at its melting point? The heat of fusion is 1.9 * 10^3 J/g.

Solution: All of the heat equations can be found on Table T in the Reference Table. An equation that involves heat of fusion is **q = mHf **where q=heat and Hf =heat of fusion. Mass and heat of fusion are given in the question.

Now, all we need to do is to plug the numbers: q=43 g *1.9 * 10^3 J/g

Answer:3

40 Solid aluminum has a specific heat capacity of 0.90 J/g•K. How many joules of heat are absorbed to raise the temperature of 24.0 grams of aluminum from 300. K to 350. K? (1) 22 J

(2) 45 J

(3) 1100 J

(4) 1200 J

Solution: All of the heat equations can be found on Table T in the Reference Table. An equation that involves specific heat capacity is** q=mCΔT** where q=heat
m = mass and C = specific heat capacity.

m=24.0 grams

C = 0.90 J/g•K

ΔT = Tfinal - Tinitial = 350K-300K = 50K

q= 24.0g* 0.90 J/g•K*50K

Answer:3

**January 2020**

13 What is the amount of heat released by 1.00 gram of liquid water at 0°C when it changes to 1.00 gram of ice at 0°C?

(1) 4.18 J

(2) 273 J (3) 334 J

(4) 2260 J

Solution: When water changes from liquid to solid phase (ice), we use Heat of fusion in our calculations. All of the heat equations can be found on Table T in the Reference Table. An equation that involves heat of fusion is **q = mHf **where *q*=heat and *Hf *=heat of fusion. Mass is 1.00 gram according to the question. Heat of fusion of water can be found in Table B and is 334 J/g. Now, we can plug everything in

q=1.00 g*334 J/g

Answer:3

**June 2019**

41 What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C? (1) 1100 J

(2) 4700 J

(3) 6300 J

(4) 11 000 J

Solution: All of the heat equations can be found on Table T in the Reference Table. An equation that involves change in temperature is **q=mCΔT **where q=heat
m = mass and C = specific heat capacity.

m=75 g from the question

ΔT = Tfinal - Tinitial = 35°C-20°C= 15°C (please note the different in temperature is the same for K and °C)

C = 4.18 J/g•K (can be found in Table B on the Reference Table)

Now, we can plug everything into the equation.

q=75 g *4.18 J/g•K *15°C

Answer:2

**January 2019**

17 The average kinetic energy of the particles in a sample of matter is expressed as (1) density

(2) volume

(3) pressure

(4) temperature

Solution: Temperature is the measure of the average kinetic energy

Answer:4

41 Which numerical setup can be used to calculate the heat energy required to completely melt 100. grams of H2O(s) at 0°C? (1) (100. g)(334 J/g) (2) (100. g)(2260 J/g) (3) (100. g)(4.18 J/g•K)(0°C)

(4) (100. g)(4.18 J/g•K)(273 K)

Solution:When water changes from solid phase (ice) to liquid, we use Heat of fusion in our calculations. All of the heat equations can be found on Table T in the Reference Table. An equation that involves heat of fusion is **q = mHf **where *q*=heat and *Hf *=heat of fusion. Mass is 100. grams according to the question. Heat of fusion of water can be found in Table B and is 334 J/g. Now, we can plug everything in

Answer:1

**August 2018**

17 A cube of iron at 20.°C is placed in contact with a cube of copper at 60.°C. Which statement describes the initial flow of heat between the cubes?

(1) Heat flows from the copper cube to the iron cube.

(2) Heat flows from the iron cube to the copper cube.

(3) Heat flows in both directions between the cubes.

(4) Heat does not flow between the cubes.

Solution: Heat always flows form an object with higher temperature to an object with lower temperature.

Answer:1

41 What is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C?

Solution:All of the heat equations can be found on Table T in the Reference Table. An equation that involves change in temperature is **q=mCΔT **where q=heat
m = mass and C = specific heat capacity. m=49.5 g from the question

ΔT = Tfinal - Tinitial = 66°C-22°C= 44°C (please note the difference in temperature is the same for K and °C)

C = 4.18 J/g•K (can be found in Table B on the Reference Table)

Now, we can plug everything into the equation and solve.

Answer:3

**January 2018**

40 Given samples of water:

Sample 1: 100. grams of water at 10.°C Sample 2: 100. grams of water at 20.°C

Compared to sample 1, sample 2 contains (1) molecules with a lower average kinetic energy

(2) molecules with a lower average velocity (3) less heat energy (4) more heat energy

Solution: Temperature is the measure of the average kinetic energy. Since the temperature of sample 2 is higher than sample 1, it would have a higher average kinetic energy, eliminating choice 1. It would, however, have more heat energy (thermal energy)

Answer:4

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