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3 years ago

What is the specific heat of a substance that absorbs 1600 joules of heat when a sample

Physics

1 answer:

kramer3 years ago

6 0

Answer:

8.08 J/g °C

Explanation:

Q=m*Cp*ΔT-->

Cp=Q/(m*ΔT) -->

Cp=1600/[18*(31-20)]-->

Cp=8.08 J/g °C

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20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solu

Anastasy [175]

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

7 0

3 years ago

A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit

Alexus [3.1K]

Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

$10^5 Q +300Q'+0.2Q''=0$

With the initial conditions as,

$Q(0) = 10^{-6}$

$Q'(0)= 0$

Where Q(t) is the charge.

<em>The general solution of a linear equation is given as:</em>

Applying this definiton in our differential equation we have that

$Q(t) = C_1e^{at}+C_2e^{bt}$

To find b and a we use the first equation and find the roots:

$r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}$

$r_{a,b} = {-1000,-500}$

Then we have

$Q(t) = C_1e^{-1000t}+C_2e^{-500t}$

To find the values of the Constant we apply the initial conditions, then

$Q(0)= 10^{-6} = C_1+C_2$

And for the derivate:

$Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}$

$0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}$

$0 = -1000C_1-500C_2$

We have a system of 2x2:

$(1) 10^{-6} = C_1+C_2$

$(2) 0 = -1000C_1-500C_2$

Solving we have:

$C_1 = -10^{-6}$

$C_2 = 2*10^{-6}$

The we can replace at the equation and we have that the Charge at any moment is given by,

$Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}$

If we obtain the derivate we find also the Current, then

$I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}$

7 0

3 years ago

An astronaut is on the moon. He drops a hammer from a height of 3.2metres and it takes 2.0 seconds to reach the lunar landscape.

Anvisha [2.4K]

Answer:

1/6 m/s^2 ( about 1/6th gravity of Earth ( 9.81 m/s^2)

Explanation:

Displacement = yo + vo t - 1/2 a t^2

- 3.2 = 0 + 0 - 1/2 a(2.0)^2

- 3.2 = -2a

a = 3.2 / 2 = 1.6 m/s^2

6 0

2 years ago

A Tale f Two Elephants

Murljashka [212]

Answer:

u need it now?

Explanation:

I'll answer it

3 0

4 years ago

solid particles or liquid drops in gas,called can stay in the upper atmosphere for months or years,reflecting away some incoming

luda_lava [24]

I'm sorry, but this is not a question. What do you need to know?

5 0

4 years ago