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QveST [7]
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?
Physics
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

Download docx
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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A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre
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Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

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q = probability of success=0.85

r=3

Formula :P(r=3)=^nC_r p^r q ^{n-r}

P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166

Standard deviation =\sqrt{n \times p \times q}

Standard deviation =\sqrt{7 \times 0.15 \times 0.85}

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

4 0
2 years ago
The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.
Vadim26 [7]

Answer:

7.9\frac{gr}{cm^3}

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7cm^3).

Density is \frac{mass}{volume}.

So the density of that piece of metal is \frac{55.3g}{7cm^3}

Which leaves us with a final density of 7.9\frac{gr}{cm^3}

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Answer:

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Explanation:

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How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated
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The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

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This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

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