Home Page Textbooks Physics Page 383: Standardized Test Prep

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 383: Standardized Test Prep

Exercise 1

Step 1

1 of 2

Energy transferred from a higher temperature object to a lower temperature object is heat.

Result

2 of 2

(A)

Exercise 2

Solution 1

Solution 2

Step 1

1 of 2

The sum of the kinetic energy and potential energy is thermal energy.

Result

2 of 2

Step 1

1 of 2

The sum of the kinetic and potential energy of an object is thermal energy. As we know that mechanical energy is the sum of the potential energy and kinetic energy and the mechanical equivalent of mechanical energy is thermal energy. The mechanical equivalent of heat is the precise amount of mechanical work that has the same effect as the transfer of a given amount of thermal energy.

The correct option is $c$ -thermal energy.

Result

2 of 2

The correct option is $c$ -thermal energy.

Exercise 3

Step 1

1 of 2

Temperature is proportional to the average kinetic energy of the particles of the object.

Result

2 of 2

Exercise 4

Step 1

1 of 3

### Knowns

– The mass of our sample $m = 1text{ kg}$

– The heating rate $P = 30 ; frac{text{J}}{text{s}}$

Step 2

2 of 3

### Calculation

To find the specific heat capacity of material A we need to find the total change in thermal energy.

From the graph, looking at the point $(T, t) = (25text{textdegree}text{C}, 60text{ s})$ we find the total time $t = 60text{ s}$ and change in temperature:

$$
begin{align*}
Delta T = T_{text{f}} – T_{text{i}} = 25text{textdegree}text{C} – 20text{textdegree}text{C} = 5text{textdegree}text{C}
end{align*}
$$

Now we can find the total thermal energy using the heating rate.

$$
begin{align*}
Q = P cdot t = 30 ;frac{text{J}}{text{s}} cdot 60text{ s} = 1800text{ J}
end{align*}
$$

Now we write the law that governs heating of objects:

$$
begin{equation*}
Q = m , c , Delta T
end{equation*}
$$

rearranging for $c$:

$$
begin{equation*}
c = frac{Q}{m , Delta T}
end{equation*}
$$

Plugging in the numbers we get:

$$
begin{align*}
c = frac{1800text{ J}}{1text{ kg} cdot 5text{textdegree}text{C}} = 360 ; frac{text{J}}{text{kg}text{textdegree}text{C}}
end{align*}
$$

Result

3 of 3

The specific heat capacity of material A is $360 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$.

So the final solution is (A) $360 ; frac{text{J}}{text{kg}text{textdegree}text{C}}$

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

If the material $A$ starts to melt at $60^circ text{C}$ then the temperature would remain constant and all the heat energy which was earlier used to raise the temperature of the material would now be used to melt the material. So if the temperature remains constant with increasing time the line on the graph will become horizontal.

The correct option is $D$.

Result

2 of 2

The correct option is $D$.

Step 1

1 of 2

Since during melting the heat energy will be used as the latent heat of melting, the temperature will remain constant at that point. So the curve will be horizontal at that point.

Result

2 of 2

(D)

Exercise 6

Solution 1

Solution 2

Step 1

1 of 2

Material $C$ has the highest average kinetic energy after $40 text{ s}$ of heating because temperature is proportional to the average kinetic energy of the particles in an object and since material $C$ has highest temperature after $40 text{ s}$ therefore it will have the highest average kinetic energy.

Thus, the correct option is $C$.

Result

2 of 2

Thus, the correct option is $C$.

Step 1

1 of 2

Since the temperature of the material C is highest after 40 s, the average kinetic energy of the molecule of material C will be highest.

Result

2 of 2

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

The results depend on the specific temperature to which the ball and ring are cooled, and thus cannot be determined. We do not know the mass of the brass ball and the brass ring and therefore we cannot determine which will change its size more, the one with smaller mass will require less change in temperature to reduce its size as compared to the other one.

Therefore, the correct option is $D$.

Result

2 of 2

Therefore, the correct option is $D$.

Step 1

1 of 2

Since both are made from same material and are of same diameter, thy will change equally with change of temperature. Hence the ball will always fit through the hole.

Result

2 of 2

(A)

Exercise 8

Solution 1

Solution 2

Step 1

1 of 2

A buildings’ heating and cooling system operates most effectively if the heating vents are located near the floor and the cooling vents are located near the ceiling because the heating vents that give out warm air which is lighter and has a tendency to rise in the upwards direction will make the room warmer faster and distribute the heat evenly, while the cold vents which give out cold air which is denser settles to the bottom, therefore if we want even cooling we must place it near the ceiling, so that cold air gets distributed evenly.

Result

2 of 2

So if we want even cooling we must place it near the ceiling, so that cold air gets distributed evenly.

Step 1

1 of 1

The heating vents heats the air surrounding it. Now if the vent is near the floor, then the heat the air near the floor. Since the hot air is light, the air will easily go up and heat the whole room.
Where as if the cooling vent is at the top, then it will cool the air at top. Now since the cold air is heavier, the cold air will easily come down and hence it will coll the room more easily.

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