Ask question

Ask question

All categories

1 month ago

9

A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the

mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C, and that of water is 1.0 cal/g ∙ C°.

1 answer:

Keith_Richards [3.2K]1 month ago

3 0

According to the details mentioned in the question, the weight of the beaker is 144.4 g.

From a complete analysis of the question;

water volume = 74 mL

Water mass = 74 g

specific heat capacity of glass = 0.18 cal/g ∙ °C

specific heat capacity of water = 1.0 cal/g ∙ C°

Mass of glass = x g

Overall heat absorbed by the system = 2000.0 cal

Increase in temperature = 20.0°C

Heat absorbed by the system = Heat absorbed by glass + Heat absorbed by water

Heat absorbed by glass = x × 0.18 × 20

Heat absorbed by water = 74 × 1.0 × 20

Therefore, it follows that

2000 = (x × 0.18 × 20) + (74 × 1.0 × 20)

2000 - 1480 = (x × 0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing sections;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C, and that of water is 1.0 cal/g ∙ C°.

Learn more:

You might be interested in

The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th

Keith_Richards [3271]

Utilizing the same generator, the only variable affecting the electric field is the speed. The higher the generator's rotation speed, the more extensive the electric field it generates. Hence, the order is 3000 rpm<3200 rpm<3400 rpm<3600 rpm

4 0

2 months ago

Read 2 more answers

Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the

Yuliya22 [3333]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

$a=\dfrac{v^2}{r}$

In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.

Thus,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}$

1 and 2 represent the first and second cars, respectively.

$v_2=2v_1$

Consequently,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1$

Therefore, car 2's acceleration is four times that of car 1.

4 0

2 months ago

Four particles with masses 2 kg, 5 kg, 2 kg, and 2 kg are connected by rigid rods of negligible mass as shown. assume the system

Maru [3345]

The answer is 10pi. I believe this will be helpful.

8 0

1 month ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

6 0

3 months ago