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tankabanditka [31]
2 years ago
5

A backpacker collects snow at 0°C, and places it in a cooking pot on a camp stove. It takes 643 kJ of heat energy to melt the sn

ow and bring the water to boiling. Assuming no heat loss, and neglecting the specific heat capacity of the pot, calculate the mass of snow that the backpacker collected. (The specific heat capacity of liquid water, c = 4.18 J/g.K; and: H₂O(s) -> H₂O(l) /\H = /\Hfusion = 6.02 kJ/mol)A) 1.92 kg B) 1.90 kg C) 1.52 kg D) 855g E) <800g
Physics
1 answer:
Nitella [24]2 years ago
7 0

Answer:

The mass of snow that the backpacker collected is: <em>D) 855g</em>

Explanation:

To boiling the water, the cooking pot requires 643kJ. This heat is due to the heat to melt the water and the heat to bring the water to boiling point.

The heat to melt the water is:

q = ΔHfusion×\frac{1mol}{18,02g}×mass

And heat to bring water to boiling point is:

q = C×mass×ΔT

Where C is specific capacity of liquid water and ΔT is change in temperature (100°C, from melting point to boling point)

As the required energy required is 643000J:

643000J =  ΔHfusion×\frac{1mol}{18,02g}×mass + C×mass×ΔT

643000J =  6020J/mol×\frac{1mol}{18,02g}×mass + 4,18J/g°C×mass×100°C

643000J = 334J/g×mass + 418J/g×mass

643000J = 752J/g×mass

855g = mass

<em>The mass of snow that the backpacker collected is 855g</em>

I hope it helps!

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On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L
vlabodo [156]

Answer:

75.6J

Explanation:

Hi!

To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,

Given the above we have the following equation.

Q=(m)(h2)-(m)(h1)

where

m=mass=1.3×10−3kg.

h2= entalpy at 37C

h1= entalpy at -20C

Q=m(h2-h1)

remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference

Q=mCp(T2-T1)

Cp= specific heat of air = 1020 J/kg⋅K

Q=(1.3×10−3)(1020)(37-(-20))=75.6J

4 0
1 year ago
In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min
Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, t_{K} = 17.9 min = 17.9\times 60 = 1074 s

Time taken by Hannah, t_{H} = 15.3 min = 15.3\times 60 = 918 s

Now, the speed of Kara and Hannah can be calculated respectively as:

v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s

v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s

Time taken in each lap is given by:

(v_{H} - v_{K})t = x

(5.45 - 4.65)\times t = 400

t = \frac{400}{0.8}

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

d_{H} = v_{H}\times t

d_{H} = 5.45\times 500 = 2725 m

No. of laps taken by Hannah when she passes Kara:

n_{H} = \frac{d_{H}}{x}

n_{H} = \frac{2725}{400} = 6.8 ≈ 7 laps

3 0
1 year ago
a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the
Katena32 [7]

Answer:

W_{earth} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 5.91 * 10^6 N : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg

g_{moon}= 1.62 \frac{m}{s^2}

g_{earth}= 9.8 \frac{m}{s^2}

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N : Rocket weight on moon

7 0
2 years ago
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mote1985 [20]

Answer:

C

Explanation:

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d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

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we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

8 0
1 year ago
As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the groun
Lera25 [3.4K]

Answer:

15m

Explanation:

Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

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the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m

X=(1.5m)(10)=15m

the nail is 15 meters above ground level

8 0
2 years ago
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