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3 years ago

10

Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 1010 j of heat. co

nvert this energy to kilocalories

1 answer:

netineya [11]3 years ago

6 0

Keeping in mind that the conversion between calories and Joules is
$1 cal = 4.186 J$
we can write the conversion factor using the kilocalories:
$1 kcal = 4186 J$

The energy released in our problem is
$E=3.3 \cdot 10^{10} J$
so we can set a simple proportion to find its equivalent in kcal:
$1 kcal: 4186 J = x: 3.3 \cdot 10^{10} J$
from which we find:
$x= \frac{3.3 \cdot 10^{10} J \cdot 1 kcal}{4186 J} =7.88 \cdot 10^6 kcal$

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"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."

These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is

"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."

The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.

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3 years ago

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Help Please I Don't Know.

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1)
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3 years ago

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba

lina2011 [118]

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T = 24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

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2 years ago

Jogger runs 400 m East in 120 s, then turns around and runs 250 m West in 90 seconds. Calculate the following values.

RSB [31]

First, we must recall that distance is the total length traveled by an object in which the direction of the motion does not matter. Meanwhile, displacement is the distance of the object from its starting point which means the directions matters for displacement.

Now, speed is distance over time while velocity is displacement over time. Since we're talking about the same object's motion, the total time traveled is (120 + 90) = 210 seconds.

Now, the total distance traveled by the object is (400 + 250) = 650 m. Meanwhile the total displacement traveled by the object is 400 m, East + 250 m, West = 150 m, East.

Now, to find the speed and velocity, we just divide the values of distance and displacement, respectively, over time. Thus, we have

speed = 650 m / 210s ≈ 3.095 m/s
velocity = 150 m, East /210 s ≈ 0.714 m/s, East

Now, rounding the values up to 2 significant digits, we have

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3 years ago

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