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3 years ago

11

Please help

1 answer:

JulsSmile [24]3 years ago

7 0

Answer:

D

Explanation:

because it is the only one that has something to do with heat keyword would be boiling

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A balloon having an initial temperature of 17.8Â°C is heated so that the volume doubles while the pressure is kept fixed. What i

Rama09 [41]

Answer:

$T = 308.6 ^0 C$

Explanation:

Here by ideal gas equation we can say

$PV = nRT$

now we know that pressure is kept constant here

so we will have

$V = \frac{nR}{P} T$

since we know that number of moles and pressure is constant here

so we have

$\frac{V_2}{V_1} = \frac{T_2}{T_1}$

now we know that initial temperature is 17.8 degree C

and finally volume is doubled

So we have

$\frac{2V}{V} = \frac{T_2}{(273 + 17.8)}$

so final temperature will be

$T_2 = 581.6 k$

$T_2 = 308.6 ^o C$

5 0

3 years ago

How much ancient plant debris is necessary to form one foot of coal

Elza [17]

Pretty sure it's 10 feet of plant debris to form 1 foot of coal.

3 0

3 years ago

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

kakasveta [241]

Answer:

$W=4037.36\ J$

Explanation:

Given:

mass of ice melted, $m=3.3\times10^{-2}\ kg$

time taken by the ice to melt, $t=5\ min=300\ s$

latent heat of the ice, $L=3.34\times 10^5\ J$

Now the heat rejected by the Carnot engine:

$Q_R=m.L$

$Q_R=0.033\times 3.34\times 10^5$

$Q_R=11022\ J$

Since we have boiling water as hot reservoir so:

$T_H=373\ K$

The cold reservoir is ice, so:

$T_L=273\ K$

Now the efficiency:

$\eta=1-\frac{T_L}{T_H}$

$\eta=1-\frac{273}{373}$

$\eta=26.81\%$

Now form the law of energy conservation:

Heat supplied:

$Q_S-W=Q_R$

where:

$Q_S=$ heat supplied to the engine

$Q_S-\eta\times Q_S=Q_R$

$Q_S(1-\eta)=Q_R$

$Q_S=\frac{11022}{1-0.2681}$

$Q_S=15059.36\ J$

Now the work done:

$W=Q_S-Q_R$

$W=15059.36-11022$

$W=4037.36\ J$

8 0

3 years ago

Simora [160]

the father has to sit 0.5meter away from the kid because he is a 3/4 heavier that the kid

6 0

2 years ago

Unpolarized light whose intensity is 1.46 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the lig

Serhud [2]

Answer:

0.73 W/m²

0.2522 W/m²

Explanation:

$I_0$ = Unpolarized light = 1.46 W/m²

$\theta$ = Analyzer angle = 54°

Light through first filter

$I_1=\frac{I_0}{2}\\\Rightarrow I_1=\frac{1.46}{2}\\\Rightarrow I_1=0.73\ W/m^2$

The intensity of the light leaving the polarizer is 0.73 W/m²

After passing through analyzer

$I=I_1cos^2\theta\\\Rightarrow I=0.73\times cos^2(54)\\\Rightarrow I=0.2522\ W/m^2$

The intensity of the light that reaches the photocell is 0.2522 W/m²

6 0

3 years ago