Ask question

Ask question

All categories

2 years ago

12

You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera

ture of 50 degrees C. The water is initially at 20 degrees C. If the equilibrium temperature of the block and water is 25 degrees C, what is the specific heat of the block? The specific heat of water is 4186 J. kg K. 3349 J, kg C 1189 J, kg C 5545 J, kg C 750 J/kg C 2080 J/kg C

1 answer:

Nina [5.8K]2 years ago

4 0

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁ = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = $\frac{2*4186*5}{0.5*25}$ = 3348.8J/kg C≅ 3349J/kg C

You might be interested in

The path a projectile takes is known as the Question 1 options: vertical component trajectory horizontal component parabola Ques

Firdavs [7]

1. Trajectory

The path a projectile is called a trajectory in physics. It has a vertical component and even makes a parabola, but if we are talking about physics, it is trajectory.

2. A person sitting in a chair

Projectiles can be defined as an object that is in flight. So it has to be in the air. Since a person sitting in a chair is not in flight, then it is NOT a projectile. (Unless you throw the person in the air while he is in the chair)

3. 490 meters

We have the formula and our given:

d = 1/2gt²

Just plug in the values to get your answer:

d = 1/2(-9.8m/s²)(10s)²

d = (-4.9m/s²)(100s²)

d = -490m

So since height is a scalar value, just take out the negative sign.

4. 65 m/s

Again we have our formula and given:

$v=\dfrac{d}{t}$

So we just plug in our values:

$v=\dfrac{650m}{10s}$

$v=650m/s$

5. True

A projectile, if you will notice its trajectory moves both horizontally and vertically. The horizontal motion is what we call the x-component and the vertical is called the y-component. This is what gives it its' curved path.

6. False

An ellipse is an oval-shaped path. A projectile does not move in a circular/oval path. It travels a curved path. It can be parabolic. It is curved but it does not follow a circular path.

7. The vertical component always equals the horizontal component

This is false. Vertical component is different from the horizontal component. The horizontal component is not influenced by gravity only the vertical component. So in short, the horizontal component is the same throughout, but the vertical component changes over time.

8. Constant

Like mentioned above, the horizontal movement is constant or it does not change. This is because a projectile is defined also as an object that is influenced solely by gravity.

9. Vertical velocity decreases

This is because the movement is against the pull of gravity. It will continue to decrease until it will eventually come to a stop and start to descend. As it descends it increases.

10. Vertical velocity increases

I guess explained it above. As the object descends, the vertical movement increases. This is why you can actually die at certain heights. It increases as the time in flight increases. So the longer in flight, the faster it will get.

3 0

3 years ago

Read 2 more answers

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0

3 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

GaryK [48]

The boat's position $x$ relative to its starting point $x_0=0$ is determined by

$x=x_0+v_0t+\dfrac12at^2$

where $v_0$ is its initial velocity, $a$ is its acceleration, and $t$ is time. After $t=6\,\mathrm s$ , the boat has traveled

$x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2$

$\implies x=91\,\mathrm m$

3 0

3 years ago

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$

8 0

3 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago