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3 years ago

How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?

Physics

1 answer:

cricket20 [7]3 years ago

7 0

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

<u>Explanation</u>:

Given, mass m = 1.25 kg, Temperature t = 100 degree celsius

To calculate the heat required,

Q = m $\times$ L

where m represents the mass in kg,

L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

Q = 1.25 $\times$ 2260

Q = 2825 kJ /kg.

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Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi

Oksanka [162]

<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>

6 0

3 years ago

What is the acceleration of an 24 kg object that applies a force of 130N?

WITCHER [35]

Answer:

To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.

Explanation:

Given ,

F = 130N

M = 24kg

A = ?

F = m× a

then ,

130N = 24kg ×a

a = 130/24 = 5 m/s.

6 0

2 years ago

Please help ASAP! Thank you :)

puteri [66]

Answer:

magnitude of gravitational force between the Earth and the Sun at B is greater than that at A

Explanation:

Formula of gravitational force:

F = GMm/r^2

(r is the distance between 2 objects)

We see that r(B) < r(A) since at B, the Earth is closer to the Sun than at A

According to the Formula, the smaller r is, the greater F is

So, F(B) > F(A)

8 0

2 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

8 0

2 years ago

A 10 µf capacitor is charged to 108 v and is then connected across a 328 ω resistor. what is the initial charge on the capacitor

valina [46]

The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
$C= \frac{Q}{V}$

The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
$Q=CV=(10 \mu F)(108 V)=1080 \mu C$

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3 years ago