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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

Tcecarenko [31]

Answer:

The amount of heat transfer is 21,000J .

Explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

Thus, the amount of heat transfer is 21,000J .

8 0

3 years ago

Read 2 more answers

The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago

A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w

Rina8888 [55]

Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
 a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)= 11.44 m/s.
 After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
 T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= 32.68 rev/s
 Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)= 1960.74 rev/min

5 0

3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s