thermal physics problem solving mr. klapholz shaker heights high school

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Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

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Page 1: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Thermal Physics

Problem Solving

Mr. KlapholzShaker Heights

High School

Page 2: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 1

If the thermal capacity of a certain amount of water is 5000 J C-1, then how much heat is needed to raise its temperature from 20 ˚C to 100 ˚C?

Page 3: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 1

C = Q / DTQ = C • DT

Q = (5000 J C-1 ) (100˚C – 20˚C)=400,000 J

= 400 kJ

Page 4: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 2

How much heat is lost from a block of metal of thermal capacity 800 J ˚C-1 when it cools from 60 ˚C to 20 ˚C?

Page 5: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 2

C = Q / DTQ = C • DT

Q = (800 J C-1 ) (20˚C – 60˚C)= -32,000 J

= -32 kJEnergy lost by a system is negative.

Page 6: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 3

How much heat do you need to add to 300 g of water to heat it from 10 ˚C to 50 ˚C? The specific heat of water is 4200 J kg-1 ˚C-1.

Page 7: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 3

Q = mc DTQ ={0.300 kg} {4200 J kg-1 ˚C-1} {50 ˚C - 10 ˚C}

Q ={300} {4200} {40} Joules= 50,400 J= 50.4 kJ

Page 8: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 4

When the temperature of a brick (1.5 kg) drops from 61 ˚C to 46˚C, the brick loses 20 kJ of heat. What is the specific heat capacity of the brick?

Page 9: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 4

Q = mc DTc = Q / { mDT }

c ={20,000 J} / {1.5 kg • 15 ˚C}c =888.9 J kg-1 ˚C-1

Page 10: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 5

The specific latent heat of fusion of water is 3.35 x 105 J kg-1. What is the minimum amount of energy is required to change 500 g of ice into water?

Page 11: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 5

Note: to find the minimum amount of energy to turn ice into liquid, only calculate the energy

required to melt the ice, not the energy required to warm the ice up to the melting temperature.

Q = LF m

Q = {3.35 x 105 J kg-1} {0.500 kg}c =1.675 x 105 J

c =167.5 kJ

Page 12: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 6

When 100 g of steam turns to liquid water, 2.27 x 105 J of energy is released. What is the specific latent heat of vaporization of water?

Page 13: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 6

Q = LV m

LV = Q / m

LV = {2.27 x 105 J } / {0.100 kg}

LV =2.27 x 106 J kg-1

Page 14: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Über Problem 7a

What is the power being delivered to the water in the first segment (bold) of the graph? The graph shows Temperature vs. Time for 1 kg of water.

Page 15: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

T / ˚C

t / s

20

60

240 480 960

100

Page 16: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 7a

Power is the rate at which energy changes form.P = Energy ÷ Time

So, for the bold segment, how much energy is delivered in how much time?

The time is 240 seconds.The energy delivered to the water is

Q = mcDTQ = (1 kg) (4200 J kg-1 ˚C-1) (60 ˚C – 20 ˚C)

Q = 168000 JP = 168000 ÷ 240 = 700 Watts = 700 J/s

Page 17: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Über Problem 7b

Starting from the when the water begins to boil, and continuing to the end of the graph, how much energy went into the water during boiling?

Page 18: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 7b

The graph shows that the water boils at 480 s, and continues to 960 s. That’s a durations of 480 s.

P = E / tE = P•t

Energy = (700 J s-1) (480 s)E = 3.36 x 105 Joules

E = 336 kJ

Page 19: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Über Problem 7c

During boiling, how much water turned to steam?

Page 20: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 7c

The energy required to vaporize a mass is:Q = Lm

m = Q ÷ Lm = {3.36 x 105 Joules} ÷ {2.27 x 106 J kg-1}

m = 0.15 kg

Page 21: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 8

A solar heater is warming some water (2 kg). The water starts out at 10 ˚C. The heater collects energy at the rate of 10 J s-1 for two hours.Set up an equation that you could use to find the final temperature of the water.

Page 22: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 8

2 hour = 7200 secondsQ = (10 J s-1) (7200 s) = 72000 J

Also, Q = mcDT7200 J = (2 kg) (4.18 x 103 J Kg-1 ˚C-1) (T2 – 10 ˚C)

From here you could solve for the final temperature.

Page 23: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Problem 10 (The Law of Mixing)This important problem shows how specific heat capacity can be measured. Put a chunk of metal (0.1 kg) in boiling water for a while, and you’ll know that the metal is at 100 ˚C. Then put the chunk into 0.4 kg of water at 10 ˚C and wait for the temperature of the water to stop rising. The equilibrium temperature is 15 ˚C. What is the heat capacity of the metal?

Page 24: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Solution 10The heat lost by the metal = the heat gained by the water.

Heat lost by chunk: -mcDT = -(0.1) c (15 - 100) = +(8.5 kg ˚C)c

Heat gained by water: mcDT = (0.4) (4180) (15 - 10) = 8360 J

(8.5 kg ˚C)c = 8360 Jc = 983 J kg-1 ˚C-1

Note: Use this method anytime a hot thing and a cold thing are put in contact. Any of the variables (m, c, T) could have been the unknown.

Page 25: Thermal Physics Problem Solving Mr. Klapholz Shaker Heights High School

Tonight’s HW:

Go through the Thermal Physics and Thermodynamics sections in your textbook and scrutinize the “Example Questions” and

solutions.Bring in your questions to tomorrow’s class.