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balandron [24]
3 years ago
7

An office worker claims that a cup of cold coffee on his table warmed up to 80°C by picking up energy from the surrounding air,

which is at 25°C. Is there any truth to his claim? Does this process violate any thermodynamic laws?
Engineering
2 answers:
Nady [450]3 years ago
6 0

Answer:

It is false, it violates the second law.

Explanation:

In order for his coffe to heat from the environment the environmnet surrounding his coffee needed to have a higher temperature than his coffee, the true claim could be that the coffee lost temperature or cooled down since the air surrounding it was cooler, the laws of thermodynamics have to do with closed systems and energy and the second states that a new object will enter a state of stabilization with the environmnet if the environment is cooler than the object ehn the object will cool off.

kherson [118]3 years ago
4 0

Answer:

The claim is false and violate the zeroth law of thermodynamics.

Explanation:

Zeroth law of thermodynamics refers to thermal equilibrium among  elements. It states that  elements which different temperatures will reach the same temperature at the endgame if they are close enough to interact each other. This temperaure is called <em>equilibrium temperature and it is always a intermediate value between the element with highest temperature and the element with the lowest one. So there is no way </em> a cup of cold coffee on a table can warm up to 80°C picking up energy from the surrounding air at 25°C because the cup can only reach a temperature closer to the surrounding air temperature which will be the equilimbrium temperature for that case.

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        The diameter of the ball is  d = 1.5 km = 1.5 *1000 = 1500m

       Hence the radius  == \frac{1500}{2} = 750m

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Where \sigma is the Stefan-Boltzmann constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}

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                    E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4

                        =1.077*10^{15} W

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                           Rate  \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4

                                                            = 5.67*10^{-8} * 7200^2

                                                            = 1.523747635*10^9 W/m^2

A sketch illustrating the b part of the question is shown on the first uploaded image

     looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

      from the diagram x can be calculated as follows

                      x = \sqrt{40^2 + 16^2}

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Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

                            G= \frac{1.077*10^15}{4 \pi (431081.3^2)}

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                            Amount \ of \ energy \ absorbed \ (E_B ) = G * t

Where t is the time taken

       Therefore     E_B = 46.177 *10 = 461.77 KJ

       

                         

                       

             

   

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