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3 years ago

Calculate the total energy transferred when 200 g of ice cubes at 0°C are changed to steam at 100°C.

Physics

1 answer:

uranmaximum [27]3 years ago

7 0

Answer:

$604000\ \text{J}$

Explanation:

m = Mass of ice = 200 g

$\Delta T$ = Temperature change of water = $(100-0)^{\circ}\text{C}$

c = Specific heat capacity of water = 4200 J/kg °C

$L_f$ = Specific latent heat of fusion = 340 kJ/kg

$L_v$ = Specific latent heat of vaporisation = 2260 kJ/kg

Heat required to convert ice to water = $mL_f$

Heat required to raise the temperature of water to boiling point = $mc\Delta T$

Heat required to convert water to steam = $mL_v$

Total heat required

$q=mL_f+mc\Delta T+mL_v\\\Rightarrow q=m(L_f+c\Delta T+L_v)\\\Rightarrow q=0.2(340\times 10^3+4200(100-0)+2260\times 10^3)\\\Rightarrow q=604000\ \text{J}$

Heat required to convert the given amount of ice to steam at the required temperature is $604000\ \text{J}$ .

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Peter left Town A at 13:30 and travelled towards Town B at an

pochemuha

Answer:

Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5

Then, the distance covered by Peter: 40x1.5= 60 miles

From 13:45 to 15:00, it pasts; 1 h 15min =1.25

Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles

Lastly, the distance between them: 60-37.5= 22.5 miles

So the answer is 22.5

4 0

4 years ago

A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust

bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0

3 years ago

While watching a planet orbiting a star, you notice that the doppler shift in the light of the star is becoming increasingly blu

Lyrx [107]

Answer:

The star must be moving "closer" to the earth because the wave fronts that are being emitted are shifted toward the blue end of the spectrum (they are closer together)

6 0

2 years ago

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

Mass of the child, m = 16 kg
Length of the incline, L = 2 m
Angle of inclination, θ = 20⁰

The vertical height of fall of the child from the top of the incline is calculated as;

$sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m$

The gravitational potential energy of the child at the top of the incline is calculated as;

$P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J$

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

2 years ago

A temperature of 200 degrees F is equivalent to approximately

Ann [662]

(1) Changing Fahrenheit to Celsius:
The formula used to convert from Fahrenheit to Celsius is as follows:
C = (F - 32) * 5/9
We are given that F=200, substitute in the above formula to get the corresponding temperature in Celsius as follows:
C = (200-32) * (5/9) = 93.333334 degrees Celsius

(2) Changing the Fahrenheit to kelvin:
The formula used to convert from Fahrenheit to kelvin is as follows:
K = (F - 32) * 5/9 + 273.15
We are given that F = 200. substitute in the above formula to get the corresponding temperature in kelvin as follows:
K = (200-32)*(5/9) + 273.15 = 366.483334 degrees kelvin

5 0

3 years ago

Read 2 more answers