Ask question

Ask question

All categories

4 years ago

10

Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec

tive for use in thermometers. If 275 J of energy are added to 0.450 kg of mercury, the mercury's temperature will increase by 4.09 K. What is the specific heat of mercury.

1 answer:

UNO [17]4 years ago

5 0

The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

$\Delta T=\frac{Q}{mC_s}$

where

$\Delta T$ is the increase in temperature

m is the mass of the sample

$C_s$ is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

$\Delta T = 4.09 K$

Therefore, by re-arranging the equation we find the mercury's specific heat:

$C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

You might be interested in

What is the circle of least confusion?

Artist 52 [7]

Answer and Explanation:

In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus while a point source is being imaged due to spherical aberration.

The Circle of Confusion is also referred to as circle of indistinctness or a blur spot

5 0

4 years ago

the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

<em>Mass of the first block, m₁ = 4.0 kg</em>
<em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

<em>T </em><em>is the tension in the connecting string due weight of the first block</em>

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

5 0

3 years ago

If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples

kvasek [131]

Answer is :

D. Quadruples

7 0

3 years ago

Read 2 more answers

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

8 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago