Home Page Textbooks Physics: Principles and Problems Page 322: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 322: Section Review

Exercise 10

Solution 1

Solution 2

Step 1

1 of 9

hfill . \
textbf{(a)} \
To convert Celsius temperature to Kelvin temperature, we use the equation below

$$
T_text{K} = T_text{C} + 273
$$

Step 2

2 of 9

hfill . \
Plugging in $T_text{C} = 5 ^circ text{C}$

$$
T_text{K} = 5 + 273
$$

$$
boxed{T_text{K} = 278 text{K}}
$$

Step 3

3 of 9

hfill . \
textbf{(b)} \
To convert Kelvin temperature to Celsius temperature, we use the equation below

$$
T_text{C} = T_text{K} – 273
$$

Step 4

4 of 9

hfill . \
Plugging in $T_text{K} = 34 text{K}$

$$
T_text{C} = 34 – 273
$$

$$
boxed{T_text{C} = -239 ^circ text{C}}
$$

Step 5

5 of 9

hfill . \
textbf{(c)} \
To convert Celsius temperature to Kelvin temperature, we use the equation below

$$
T_text{K} = T_text{C} + 273
$$

Step 6

6 of 9

hfill . \
Plugging in $T_text{C} = 212 ^circ text{C}$

$$
T_text{K} = 212 + 273
$$

$$
boxed{T_text{K} = 485 text{K}}
$$

Step 7

7 of 9

hfill . \
textbf{(d)} \
To convert Kelvin temperature to Celsius temperature, we use the equation below

$$
T_text{C} = T_text{K} – 273
$$

Step 8

8 of 9

hfill . \
Plugging in $T_text{K} = 316 text{K}$

$$
T_text{C} = 316 – 273
$$

$$
boxed{T_text{C} = 43 ^circ text{C}}
$$

Result

9 of 9

(a) $T_text{K} = 278 text{K}$

(b) $T_text{C} = -239 ^circ text{C}$

(d) $T_text{C} = 43 ^circ text{C}$

Step 1

1 of 2

$T_K = T_C + 273 = 5 + 273 = 278 K$

$T_C = T_K – 273 = 34 – 273 = -239 text{textdegree} C$

$T_K = T_C + 273 = 212 + 273 = 485 K$

$T_C = T_K – 273 = 316 – 273 = 43 text{textdegree} C$

Result

2 of 2

a) $278 K$

b) $-239 text{textdegree} C$

c) $485 K$

d) $43 text{textdegree} C$

Exercise 11

Solution 1

Solution 2

Step 1

1 of 7

hfill . \
To convert Celsius temperature to Kelvin temperature, we use the following equation

$$
T_text{K} = T_text{C} + 273
$$

Step 2

2 of 7

hfill . \
textbf{(a)} \
Plugging in $T_text{C} = 28 ^circ C$ to Equation (1)

$$
T_text{K} = 28 + 273
$$

$$
boxed{T_text{K} = 301 text{K}}
$$

Step 3

3 of 7

hfill . \
textbf{(b)} \
Plugging in $T_text{C} = 28 ^circ C$ to Equation (1)

$$
T_text{K} = 154 + 273
$$

$$
boxed{T_text{K} = 427 text{K}}
$$

Step 4

4 of 7

hfill . \
textbf{(c)} \
Plugging in $T_text{C} = 568 ^circ C$ to Equation (1)

$$
T_text{K} = 568 + 273
$$

$$
boxed{T_text{K} = 841 text{K}}
$$

Step 5

5 of 7

hfill . \
textbf{(d)} \
Plugging in $T_text{C} = -55 ^circ C$ to Equation (1)

$$
T_text{K} = (-55) + 273
$$

$$
boxed{T_text{K} = 218 text{K}}
$$

Step 6

6 of 7

hfill . \
textbf{(e)} \
Plugging in $T_text{C} = -184 ^circ C$ to Equation (1)

$$
T_text{K} = (-184) + 273
$$

$$
boxed{T_text{K} = 89 text{K}}
$$

Result

7 of 7

(a) $T_text{K} = 301 text{K}$

(b) $T_text{K} = 427 text{K}$

(d) $T_text{K} = 218 text{K}$

(e) $T_text{K} = 89 text{K}$

Step 1

1 of 2

a) $T_K = T_C + 273 = 28 + 273 = 301 K$

b) $T_K = T_C + 273 = 154 + 273 = 427 K$

c) $T_K = T_C + 273 = 568 + 273 = 841 K$

d) $T_K = T_C + 273 = (-55) + 273 = 218 K$

e) $T_K = T_C + 273 = (-184) + 273 = 89 K$

Result

2 of 2

a) $301 K$

b) $427 K$

c) $841 K$

d) $218 K$

e) $89 K$

Exercise 12

Solution 1

Solution 2

Step 1

1 of 1

Thermal energy is total energy of all the molecules in a single object. Temperature measures how much energy there is per molecule. If you have two of the same bowls with the same amount of water in them, they will have the same number of molecules. The bowl that has hot water ill have more thermal energy than the bowl with cold water.

Step 1

1 of 1

The thermal energy is the kinetic energy of all the molecules of a substance. Since temperature is directly proportional to the kinetic energy of molecules, if temperature is larger the thermal energy of a substance will also be larger, so the conclusion is that if there is a same amount of water in both bowls the bowl with hot water will have larger thermal energy.

Exercise 13

Solution 1

Solution 2

Step 1

1 of 2

The baked potato stays hot longer than any other food because it has high heat capacity. This means that the heat required to change its temperature is greater compared to objects with low heat capacity.

Result

2 of 2

See the explanation

Step 1

1 of 1

The baked potato has a large specific heat compared to other foods. Objects that have a high specific heat will conduct heat poorly which means that it will lose heat energy slower than objects that have low specific heat.

Exercise 14

Solution 1

Solution 2

Step 1

1 of 1

The floor tiles are at the same temperature as the rest of the room; however, the floor tiles could conduct heat easier than other materials. This means that the floor could conduct heat from someone’s feet which makes the person feet feel cooler.

Step 1

1 of 1

Our bodies are not great at detecting the temperature of an object and the reason why some object feels colder even though it is in state of thermal equilibrium is because the heat is being transferred from our bodies to that object and that is what we feel, but for that to happen the object must be a good heat conductor and floor tiles usually are.

Exercise 15

Solution 1

Solution 2

Step 1

1 of 1

The plastic spoon must have a lower specific heat than the hot cocoa. This will mean that the plastic spoon will not transmit that much heat to your tongue as it cools down.

Step 1

1 of 1

If the spoon was in the cocoa long enough that it reached the same temperature as the hot cocoa, then, when you put it on your tongue it does not burn because plastic is a bad heat conductor i.e., it has low specific heat capacity so it slowly transmits heat to your tongue as it cools.

Exercise 16

Solution 1

Solution 2

Step 1

1 of 1

Thick aluminum is better for cooking because it conducts heat much better. Also, thick aluminum doesn’t have hot spots like the thin aluminum does.

Step 1

1 of 1

Since the heat capacity is proportional to the mass of the object if the aluminum is thicker it slowly absorbs and slowly releases heat which makes it better for cooking as it creates a more controlled environment.

In contrast, thin aluminum heats up faster and releases heat faster so the food is more susceptible to burning.

Exercise 17

Solution 1

Solution 2

Step 1

1 of 1

Potatoes are great heat conductors. If we cut the potato into small pieces, you can increase it’s surface area and allow heat flow throughout the potato.

Step 1

1 of 1

Imagine that we want to heat something up, we take the object we want to heat up and put it in contact with the warmer object. After the contact is made the molecules from the warmer object transfer their energy to the molecules of the colder object and the molecules in the colder object then redistribute that energy to other molecules around them. If more molecules of the colder object are brought into contact with the molecules of the hotter object heating will be faster. In other words, if the contact surface between the colder and warmer object is greater the heating will be better.

Now consider a whole uncut potato. It has a small contact surface with the pan it is in, so in order to cook the potato faster we cut it up in a way so that it has greater contact surface with the pan and that usually means cutting it up into french fries. The contact surface gets even better if we put the potato into some liquid so we put the french fries into the oil that conducts heat very well.

Exercise 18

Step 1

1 of 2

The coolest part of the water in the boiling pot is at the bottom. The increase in the thermal energy of the water molecules causes them to rise to the surface until they eventually escape the liquid and transform into mist/steam.

Result

2 of 2

See the explanation

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Page 322: Section Review

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