Physics, 23.11.2020 21:40 jay8682

Q18: A cube of aluminum has an edge length of 20 cm. Aluminum has a density of 2.7 g/cm and a specific heat of 0.217 cal/ g.°С. When
the internal energy of the cube increases by 47000 cal its temperature
increases by:
A
B
C
D
E
5 °C
10 °C
20 °C
100 °C
200 °C

Answers: 2

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Answers

Answer from: ashton100426owz1ly

The change in temperature of this cube of aluminum is equal to: B. 10°C

Given the following data:

Edge length, L = 20 cm.Density of Aluminum = 2.7 g/cmSpecific heat capacity (C) of aluminum = 0.217 Cal/g°СInternal energy = 47000 calories.

To find the change in temperature of this cube of aluminum:

First of all, we would determine the volume of this cube of aluminum.

$Volume \;of \;a \;cube = L^3\\\\Volume \;of \;a \;cube = 20^3\\\\Volume \;of \;a \;cube = 8000\; cm^3$

Next, we calculate the mass of this cube of aluminum:

$Mass = Density \times Volume\\\\Mass = 2.7 \times 8000$

Mass = 21,600 grams.

Now, we can find the change in temperature of this cube of aluminum:

Mathematically, the quantity of heat energy is given by the formula;

$Q = mc\theta$

Where:

Q represents the quantity of heat energy. m represents the mass of an object. c is the specific heat capacity. ∅ is the change in temperature.

Substituting the parameters into the formula, we have;

$47000 = 21600 \times 0.217 \times \theta\\\\47000 = 4687.2 \theta\\\\ \theta =\frac{47000}{4687.2} \\\\ \theta = 10.03$

Change in temperature = 10°C

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Answer from: dbanks701

10 °C

Explanation:

From the question given above, the following data were obtained:

Egde length (L) of aluminum = 20 cm

Density of Aluminum = 2.7 g/cm³

Specific heat capacity (C) of aluminum = 0.217 cal/ g°С

Heat (Q) energy = 47000 cal

Change in Temperature (ΔT) =?

Next, we shall determine the volume of the aluminum. This can be obtained as follow:

Egde length (L) of aluminum = 20 cm

Volume (V) of aluminum =?

V = L³

V = 20³

V = 8000 cm³

Thus, the volume of the aluminum is 8000 cm³

Next, we shall determine the mass of the aluminum. This can be obtained as follow:

Density of Aluminum = 2.7 g/cm³

Volume of Aluminum = 8000 cm³

Mass of aluminum =.?

Density = mass/volume

2.7 = mass /8000

Cross multiply

Mass of aluminum = 2.7 × 8000

Mass of Aluminum = 21600 g

Finally, we shall determine the change in temperature of the aluminum as follow:

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Heat (Q) energy = 47000 Cal

Mass (M) of Aluminum = 21600 g

Change in Temperature (ΔT) =?

Q = MCΔT

47000 = 21600 × 0.217 × ΔT

47000 = 4687.2 × ΔT

Divide both side by 4687.2

ΔT = 47000 / 4687.2

ΔT = 10 °C

Therefore, the increase in the temperature of the aluminum is 10 °C.

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Q18: A cube of aluminum has an edge length of 20 cm. Aluminum has a density of 2.7 g/cm and a speci...

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