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How much energy is needed to melt 150 g of ice at 0°C to water? (1)(Lf =3.34˟ 10⁵ J/Kg)

NARA [144]

Answer:

5.01×10⁴ J.

Explanation:

Applying,

q = Cm....................... Equation 1

Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.

From the question,

Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg

Substitute these values into equation 1

q = (0.15×3.34×10⁵)

q = 0.501×10⁵ J

q = 5.01×10⁴ J.

5 0

3 years ago

An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the

Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

u=0m/s t=1 s a=9.8m/s²

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

u=0m/s t=2 s a=9.8m/s²

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

u=0m/s t=3 s a=9.8m/s²

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

u=0m/s t=4 s a=9.8m/s²

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

u=0m/s t=5 s a=9.8m/s²

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0

2 years ago

A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig

Blababa [14]

Answer:

Please Mark As Brainliest!!

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a) If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)

W = 7.03 J

4 0

3 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0

3 years ago

What is the conclusion of coin and feather experiment?

valina [46]

Answer:

So the conclusion is that in presence of air net force acting downward reduces for feather and hence falls slower than coin. But in absence of air resistance, net downward force is just equal to force due to gravity which is same for both coin and feather and hence they fall down at the same rate.

5 0

3 years ago