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2 years ago

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The metal with the highest melting temperature is tungsten which melts at around 3400 K. What is the wavelength of the peak of t

he blackbody distribution for this temperature. Is this visible, infrared or ultraviolet?

1 answer:

alekssr [168]2 years ago

7 0

Explanation:

It is given that, the metal with the highest melting temperature is tungsten which melts at around 3400 K, T = 3400 K

We need to find the wavelength of the peak of the black body distribution for this temperature. It can be calculated using Wein's displacement law as :

$\lambda\times T=k$

k is the constant, $k=2.89\times 10^{-3}\ m.k$

$\lambda=\dfrac{k}{T}$

$\lambda=\dfrac{2.89\times 10^{-3}\ m.k}{3400\ k}$

$\lambda=8.5\times 10^{-7}\ m$

or

$\lambda=850\ nm$

The wavelength of infrared is from 700 nm to 1 mm. So, the lies in infrared region of the spectrum. Hence, this is the required solution.

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How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the

GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of steam from 125.5 °C to 100 °C is:

$Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J$

The heat required to change the steam at 100°C to water at 100°C is;

$Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J$

The heat required to change the temperature from 100°C to 0°C is

$Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J$

The heat required to change the water at 0°C to ice at 0°C is:

$Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J$

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

$Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C) \\ \\ Q_5 = 7132.125 \ J$

The total heat required to change the steam into ice is:

$Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J$

b)

The time taken to convert steam from 125 °C to 100°C is:

$t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W} = 8.12 \ s$

The time taken to convert steam at 100°C to water at 100°C is:

$t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s$

The time taken to convert water to 100° C to 0° C is:

$t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s$

The time taken to convert water at 0° to ice at 0° C is :

$t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08 \ s$

The time taken to convert ice from 0° C to -19.5° C is:

$t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55 \ s$

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2 years ago

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2 years ago

A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a

tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

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

where:

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the engine in the problem, we know that

$\eta = 0.400$ is the efficiency

$T_C = 298.0 K$ is the temperature of the cold reservoir

Solving for $T_H$ , we find:

$\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K$

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2 years ago

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

Bond [772]

Answer:

$S_{s}=300 m/s$

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Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

$\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s$

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

$\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km$

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0

2 years ago

A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0

2 years ago