All categories

3 years ago

How much energy is required to take 500g of ice at -40 degrees to water at room temperature (22 degrees Celsius)? Show work pls

Physics

1 answer:

Minchanka [31]3 years ago

3 0

Answer: Given that:

Mass (m) = 20 g = 0.02 Kg,

temperature (T₁) = -10°C = -10+273 = 263 K

temperature (T₂) = 10°C = 10+273 = 283 K

Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)

= 0.02 × 4.187 × (283-263)

Q = 1.67 KJ

Heat transferred is 1.67 KJ

Explanation:

You might be interested in

a child is stationary on a swing. The child is given a push by a parent and the child starts swinging

Nesterboy [21]

Answer:

you havent given the full question

but im guessing momentum

momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity or the impetus gained by a moving object.

Explanation:

as the child is pushed, it gathers momentum as its weight allows it be pushed forward, and the velocity is the speed driven by the amount of force the parent pushes on the child whilst they are swinging. The momentum is the result of this action

the equation that links these factors together are

p = mv

p = momentum

m = mass

v = velocity

hope i got it right ._.

3 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$