Ask question

Ask question

All categories

3 years ago

14

Brass is an alloy made from copper and zinc a 0.59 kg brass sample at 98.0 is dropped into 2.80 kg of water at 5.0 c if the equi

librium temperature is 6.8 what is the specific heat capacity

2 answers:

MariettaO [177]3 years ago

6 0

Answer:

393.399 J/kg.°C

Explanation:

Specific heat capacity: This is the quantity of heat required to raise the temperature of a unit mass of a substance through a degree rise in temperature.

Heat lost by the brass = heat gained by water

CM(t₁-t₃) = cm(t₃-t₂)........................ Equation 1

Where C = specific heat capacity of the brass, M = mass of the brass, t₁ = initial temperature of the brass, t₂ = initial temperature of water, t₃ = temperature of the mixture.

Making C the subject of the equation

C = cm(t₃-t₂)/M(t₁-t₃)............................... Equation 2

Given: M = 0.59 kg, m = 2.8 kg, t₁ = 98 °C, t₂ = 5.0 °C, t₃ = 6.8 °C

Constant: c = 4200 J/kg.°C

Substitute into equation 2,

C = 2.8×4200(6.8-5.0)/0.59(98-6.8)

C = 21168/53.808

C = 393.399 J/kg.°C

Thus the specific heat capacity of the brass = 393.399 J/kg.°C

zhannawk [14.2K]3 years ago

3 0

The specific heat capacity of brass would be ranked between 0 and infinity

You might be interested in

A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m

Alina [70]

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

$2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}$

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

$K=\frac{1}{2}mv^{2}$

K = 0.5 x 3 x 3 x 3

K = 13.5 J

8 0

4 years ago

A nerve signal travels 150 meters per second. Determine the number of kilometers that the nerve signal will travel in the same t

joja [24]

Explanation:

It is given that,

A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.

We know that,

1 kilometer = 1000 meter

1 hour = 3600 seconds

$150\ \dfrac{m}{s}=150\times \dfrac{1/1000\ km}{1/3600\ h}$

$150\ m/s=540\ km/h$

So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.

6 0

3 years ago

awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall

pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

$KE = 6.66 J$

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

$m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}$

now plug in all values in it

$1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22$

so we have

$3.01 = 0.375v_1 + 0.927v_2$

similarly in Y direction we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}$

now plug in all values in it

$0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22$

so we have

$0 = 0.927v_1 - 0.375v_2$

$v_2 = 2.47 v_1$

now from 1st equation we have

$3.01 = 0.375 v_1 + 0.927(2.47 v_1)$

$v_1 = 1.13 m/s$

$v_2 = 2.78 m/s$

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

$KE = \frac{1}{2}(1.47)(3.01^2)$

$KE = 6.66 J$

5 0

4 years ago

A car starts from rest with an acceleration of 6

mrs_skeptik [129]

Answer:

$t=16.67s$

Explanation:

From the question, acceleration from $6m/s^2 \ to \ 0m/s^2$ in 10 seconds.

Acceleration function can be written as:-

$a_t=6-0.6t$

From the acceleration equation, we can obtain the velocity equation:

$v(t)=\int\limits^t_b {0} \, dv=\int\limits^t_0 {a(t)} \, dt\\v(t)=\int\limits^t_0 {(6-06t)} \, dt=6t-0.3t^2\\*dv=a(t)dt$

We calculate velocity after 10sec from the above v(t) as $30m/s$ .

To obtain distance travelled after 10 seconds:-

$S=\int\limits^{10}_0 {6t-0.3t^2} \, dt=|3t^2-0.1t^3| \ *lims(10,0)\\ S=200m$

Therefore 200m is for 10seconds, and next 200m at 30m/s

Total time= $10+\frac{200}{30}=16.67s$

4 0

4 years ago

How does vapor release energy into the atmosphere

Oduvanchick [21]

I would say a because i just had this

4 0

3 years ago

Read 2 more answers