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4 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]4 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]4 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

Why do things become hot or cold?<br> Relate to Thermal Energy and The Law of Conservation of Energy

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<u>Answer:</u>

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<u>Explanation:</u>

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3 years ago

What ia the law of convesation of energy

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4 years ago

A body on the surface of a planet with the same radius as Earths weighs 10 times more than it does on Earth. What is the mass of

nadya68 [22]

Answer:

Explanation:

Given

radius of Planet is equal to radius of Earth

$r_p=r_e$

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where m=mass of body

$g_p=acceleration\ due\ to\ gravity\ on\ surface\ of\ Planet$

Weight of body on earth $F_e=mg_e$

$g_e=acceleration\ due\ to\ gravity\ on\ Earth$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

where G=gravitational constant

M=mass of Planet

r=radius of planet

for earth $g_e=\frac{GM_e}{r_e^2}$

for planet $g_p=\frac{GM_p}{r_p^2}$

substituting these values in $F_e$ and $F_p$

$F_p=m\times \frac{GM_p}{r_p^2}---1$

$F_e=m\times \frac{GM_e}{r_e^2}---2$

divide 1 and 2

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3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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3 years ago