WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE QUESTION!

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

When the energy of a sound wave is transferred to a particle of a medium and causes it to vibrate and generate heat the sound wa

skelet666 [1.2K]

The answer of this said question is d

3 0

3 years ago

If the mass of a wheel is increased by a factor of 7 and the radius is increased by a factor of 15, by what factor is the moment

inn [45]

Answer:

The moment of inertia will be increased by a factor of 1575

Explanation:

Let the moment of inertia of a wheel be given as;

$I = \frac{1}{2} mr^2\\\\$

where;

I is the moment of inertia of the wheel

m is mass of the wheel

r is radius of the wheel

when the mass is increased by a factor of 7 and the radius is increased by a factor of 15

$I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 = \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1$

Thus, the moment of inertia will be increased by a factor of 1575

3 0

3 years ago

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi

k0ka [10]

Answer:

a) F₁₂₀ = 1.34 pa A , b) F₂₀ = 0.746 pa A

Explanation:

Part. A . The definition of pressure is

P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

P V = n R T

Let's write this equation for two temperatures

P₁ V = n R T₁

P₂2 V = n R T₂

P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

P₂ = P₁ T₂ / T₁

P₂ = pa 393/293

P₂ = 1.34 pa

We calculate the strength

P₂ = F₁₂₀ / A

F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

P₁ / P₂ = T₁ / T₂

P₂ = P₁ T₂ / T₁

P₂ = pa 293/393

P₂ = 0.746 pa

Let's calculate the force (F20), from this point

F₂₀ / A = 0.746 pa

F₂₀ = 0.746 pa A

6 0

3 years ago

1) The smallest part of an element that behaves like the element is the atom.

Mandarinka [93]

Here are the answers:

1. False - Molecules is the smallest part of an element that behaves like the element.

2. False - The nucleus contains both protons and neutrons

3. True

4. True

5. A. Nucleus

6. D. Neutron

7. B. Protons and Neutrons

8. C. Electron

9. C. 6

10. C.6

5 0

4 years ago