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Allisa [31]
3 years ago
11

What temperature increase is necessary to increase the power radiated from an object by a factor of 8?

Physics
1 answer:
Aleks04 [339]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

P = \sigma T^4

Making the relationship for two states we have to

\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}

Since the final power is 8 times the initial power then

P_2 = 8P_1

Substituting,

\frac{1}{8} = \frac{T_1^4}{T_2^4}

T_2 = T_1 8*(\frac{1}{4})

T_2 = 1,68T_1

The temperature increase would then be subject to

\Delta T = T_2-T_1

\Delta T = 1.68T_1 -T_1

\Delta T = 0.68T_1

The correct option is D, about 68%

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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos
Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh

there, no vertical movement

So, on neglect of height term

P_{1}=P+\dfrac{1}{2}\rho v_{1}^2

Where, P= atmospheric pressure

\rho = air density

v = speed

Put the value in the equation

P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)

P_{1}=2126.0\ lb/ft^2

P_{1}=\dfrac{2126.0}{144}

P_{1}= 14.76\ psi

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{2}=P+\dfrac{1}{2}\rho v_{2}^2

Put the value in the equation

P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)

P_{2}=2246.17\ lb/ft^2

P_{2}=\dfrac{2246.17}{144}

P_{2}= 15.59\ psi

(c).  Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{3}=P+\dfrac{1}{2}\rho v_{3}^2

Put the value in the equation

P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)

P_{3}=2257.93\ lb/ft^2

P_{3}=\dfrac{2257.93}{144}

P_{3}= 15.68\ psi

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

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3 years ago
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