Ask question

Ask question

All categories

2 years ago

11

1. A liquid of mass 250g is heated with an electric heater. Its temperature rises from 30°C to 80°C, the specific heat capacity

of the liquid is 2000J/kg°C and the power rating of the heater is 500W. For how long was the heater used?
2. A solid metal of mass 2kg is heated using a heater of power 2kW for 5 minutes. Its final temperature becomes 400°C. Find the initial temperature of the metal. Specific heat capacity of the metal=1J/g°C.

1 answer:

statuscvo [17]2 years ago

8 0

Answer:

1) 50 seconds 2) 100°C

Explanation:

(Follows formula of Power=Energy/Time)

1) 500W x X = 2000J/kg°C x .25kg x 50°C

X = 50 seconds.

2) 2000W x 300s = 1000J/kg°C x 2kg x X

X = 300

Initial temperature => 400°C-300°C = 100°C

You might be interested in

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as $x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

$\omega =\sqrt{\frac{k}{m}}$ is the natural frequency of the system

At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

4 0

1 year ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

2 years ago

olga nikolaevna [1]

As he lifts the sack to his chest from the floor

5 0

2 years ago

Read 2 more answers

Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr

svlad2 [7]

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, $\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2$

Surface charge density of second surface, $\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2$

The electric field at a point between the two surfaces is given by :

$E=\dfrac{\sigma}{2\epsilon_o}$

$E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}$

$E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}$

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

6 0

2 years ago

A rural mail carrier is driving slowly, putting mail in mailboxes near the road

kow [346]

Incomplete question. I inferred you want access to the full question.

Explanation:

A rural mail carrier is driving slowly, putting mail in mailboxes near the road. He overshoots one mailbox, stops, shifts into reverse, and then backs up until he is at the right spot. The velocity graph given below represents his motion.

a. draw the mail carrier's position vs time graph. Assume that x = 0 m at t = 0 s.

b. what is the position of the mailbox?

4 0

2 years ago