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3 years ago

12

Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem

perature. Th is is one of the reasons why desert regions, although very hot during the day, are bitterly cold at night. Th e heat capacity of air at room temperature and pressure is approximately 21 J K−1 mol−1. How much energy is required to raise the temperature of a room of dimensions 5.5 m × 6.5 m × 3.0 m by 10°C? If losses are neglected, how long will it take a heater rated at 1.5 kW to achieve that increase given that 1 W = 1 J s−1?

2 answers:

LenKa [72]3 years ago

6 0

Answer:

The energy is 2.201x10⁴kJ and the time is 1.467x10⁴ s

Explanation:

please, the solution is in the attached Word file

Darina [25.2K]3 years ago

4 0

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

$V= 5.5\times 6.5\times 3\ m^3$

$V=107.25\ m^3$

$V=107.25\times 1000 L$

V= 107250 L

At STP number of moles given as

$n=\dfrac{V}{V_{at\ S.T.P.}}$

V=22.4 L at S.T.P.

$n=\dfrac{107250}{22.4}\ moles$

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

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3 years ago

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3 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

2 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago