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3 years ago

How many calories are released when 6 grams of 100°C steam turns to 0°C ice?

Physics

1 answer:

maks197457 [2]3 years ago

6 0

Answer: 4276.2 calories

Explanation:

Given

mass of steam is 6 gm at $100^{\circ}C$

Conversion of steam to ice involves

steam to water at $100^{\circ}C$

water at $100^{\circ}C$ to water at $0^{\circ}C$

water to the ice at $0^{\circ}C$

Calories released during the conversion of steam to water at $100^{\circ}C$

$E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6\times 533=3198\ cal.$

Calories released during the conversion of water at $100^{\circ}C$ to water at $0^{\circ}C$

$E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^{\circ}C]\\E_2=6\times 1\times 100=600\ cal.$

Calories released during the conversion of water to the ice at $0^{\circ}C$

$E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6\times 79.7=478.2\ cal.$

The total energy released is

$E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.$

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You whirl a stone in a horizontal circle in such a way that the stone is in uniform circular motion. Which of the following is t

horsena [70]

a. The direction of the stone's velocity changes as it moves around the circle.

b. The magnitude of the stone's velocity does not change.

d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.

Above given are true for the given situation.

<u>Answer:</u> Option A, B and D

<u>Explanation:</u>

Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.

The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.

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3 years ago

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms2.75 ms from an init

kumpel [21]

Answer:

2677.3 N

Explanation:

v₀ = initial speed of the hand = 4.75 m/s

v = final speed of the hand = 0 m/s

m = Total mass of hand and forearm = 1.55 kg

t = time interval for hand to come to rest = 2.75 ms = 0.00275 s

F = Force applied on the leg

Using Impulse-change in momentum equation

F t = m (v - v₀)

F (0.00275) = (1.55) (0 - 4.75)

F = - 2677.3 N

magnitude of force = 2677.3 N

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3 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

a cart has 30J or potential energy at the top of a 20m high hill. Assuming no energy is transformed to another form of energy, h

topjm [15]

Answer:

i would say 20J

Explanation:

3 0

3 years ago

The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

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3 years ago