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andreev551 [17]
3 years ago
11

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

ime interval sunshine coming through the windows delivered 2000 J of energy into the house. The temperature of the house didn't change. What was ?Ethermal of the house, and what was |Q|, the magnitude of the energy transfer between the house and the outside air?
?Ethermal = J

|Q| = J
Physics
1 answer:
Bumek [7]3 years ago
5 0

Explanation:

Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.

Hence, \Delta E = 0. And, as \Delta E = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

                   Q = -(19000 J + 2000 J)

                       = -21000 J

or,    |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

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A stone is dropped from a cliff. What will be its speed when it was fallen 100 m?
Mars2501 [29]

Answer:

final velocity will be44.72m/s

Explanation:

HEIGHT=h=100m

vi=0m/s

vf=?

g=10m/s²

by using third equation of motion for bodies under gravity

2gh=(vf)²-(vi)²

evaluating the formula

2(10m/s²)(100m)=vf²-(0m/s)²

2000m²/s²=vf²

√2000m²/s²=√vf²

44.72m/s=vf

6 0
3 years ago
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Two different liquids are poured into a jar until it is half full. The jar is then sealed shut and shaken. The liquids undergo a
Lilit [14]

Answer:

A closed system.

Explanation:

The three major types of system are: open, closed and isolated. Open system interacts with its surroundings with respect to its particles and energy. A closed system interacts with its surroundings with respect to energy but not its particles. While an isolated system does not interact with its surroundings in any way.

Therefore, after the jar is sealed, it is an example of a closed system. This is because the emitted gas could not escape into the surroundings, but thermal energy was emitted into its surroundings after the chemical reaction has taken place.

7 0
3 years ago
A non-_____ rock has interlocking grains with no specific pattern.
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A non <span>foliated </span>rock has interlocking grains with no specific pattern.
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4 years ago
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Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
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