Ask question

Ask question

All categories

2 years ago

8

An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

ecific heat of the material? (Unit=J(kg*C))

1 answer:

Sergeeva-Olga [200]2 years ago

4 0

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = <u>7</u><u>3</u><u>5</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

You might be interested in

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

olga2289 [7]

The conclusion that is best supported by the data is;

D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

3 0

2 years ago

Read 2 more answers

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

2 years ago

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener

AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0

2 years ago

A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi

Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

$R=\frac{u^{2}Sin2\theta }{g}$

$R=\frac{381^{2}Sin147 }{9.8}$

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0

3 years ago

A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)

Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s$

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0

2 years ago