All categories

4 years ago

What is the change in internal energy if 60 J of heat are released from a

Physics

1 answer:

lana [24]4 years ago

7 0

Answer:

60 J of heat are released from the system: Q=-60 J

20 J of work is done on the system: W=-20 J

∆U=Q-W

∆U=-60-(-20)

∆U=-40 J

Explanation:

If the heat is released from the system, Q should be negative.

If the heat is gained by the system, Q should be positive.

If the work is done on the system, W should be negative.

If the work is done by the system, W should be positive.

You might be interested in

Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

Mariana [72]

a) Density of the liquid: $823.5kg/m^3$

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate:

a) the density of the liquid;

b) the weight of the liquid."

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

$M=m_L + m_V$ (1)

where

$m_L$ is the mass of the liquid

$m_V$ is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

$M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg$

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

$m_L=M-m_V=175-35=140 kg$

The density of the liquid is given by

$d=\frac{m}{V}$

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = $0.170 m^3$ (volume of the liquid, which is equal to the volume of the vessel)

So we get

$d=\frac{140}{0.170}=823.5kg/m^3$

The weight of a body is given by

$F=mg$

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

$g=9.8 m/s^2$ (acceleration due to gravity)

Therefore, its weight is

$F=(140)(9.8)=1372 N$

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

6 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$