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Alexxandr [17]
3 years ago
10

You're sitting on a warm granite rock, enjoying the sunshine. You decide it's time to test the water. You take off your shoes an

d walk across the cool sand to the water. You wade into the water, enough to cover your feet. Brrr! The water is cold. You will have to wear a dry suit when you go diving.
a. Which item - the granite rock, the sand, or the water - has the highest specific heat? Which has the lowest specific heat?
b. Which item requires the least energy to increase its temperature? Explain your answer in terms of this item's specific heat.
c. If the granite rock on which you were sitting has a specific heat of 600 J/(kg°C), how much energy must be added to a 2-kilogram piece of it to increase its temperature from 20°C to 30°C? Show your work, including the equation you need to perform this calculation.
d. What is the second law of thermodynamics? Use this law to describe why your feet got cold, not hot, when you put them in the water.
Physics
1 answer:
Morgarella [4.7K]3 years ago
4 0

Answer: seen below

Explanation:

a) Water has the highest specific heat capacity in this case while sand has the lowest specific heat capacity.

b) Sand absorb the least energy for it temperature change (increase or decrease). This means that their temperatures change more quickly due to it low specific heat capacity of 0.290J/g°C) at 25°C)

c) using the formula:

Q = m × cP × (final temperature -initial temperature)

where:

Q = heat

m = is the mass

cP = is the mass heat capacity,

T= is the temperature

Q = 2 kg × 600 J/(kg°C) × (30°C - 20°C)

Q = 12000 J

d) in simpler term, The second law of thermodynamics states that when energy changes from one form to another form, entropy The second law of thermodynamics says that when energy changes from one form to another form, or matter moves freely, entropy (which is a measure of the amount of energy in a physical system that cannot be used to do work) in a closed system increases. This is why your feet gets cold when you put them in the water.

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julia-pushkina [17]

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<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

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3 years ago
Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
koban [17]

Answer:

A

   N  = 1340.86 \ slits  / cm

B

    \theta  = 15.7^o

Explanation:

From the question we are told that  

      The wavelength is  \lambda  =  650 \  nm  =  650  *10^{-9} \  m  

        The angle of  first bright fringe is  \theta  =  5^o  

        The order of the fringe considered is  n  =1

Generally the condition for constructive interference is  

       dsin (\theta ) = n * \lambda

=>    d =  \frac{1 *  650 *10^{-9 }}{ sin(5)}

=>    d = 7.458 *10^{-6} \  m

Converting to cm

           d = 7.458 *10^{-6} \  m = 7.458 *10^{-6}  * 100 =  0.0007458 \  cm

Generally the number of grating pre centimeter is  mathematically represented as

           N  =  \frac{1}{d}

=>         N  =  \frac{1}{0.0007458}

=>         N  = 1340.86 \ slits  / cm

Considering question B  

   From the question we are told that

     The first wavelength is  \lambda_1 =  650 \ nm  =  650 *10^{-9} \  m

     The second wavelength is  \lambda_2 = 429 \  m   =   420 *10^{-9 } \  m

      The order of the fringe is  n  =  2

       The grating is  N =  5000 \  slits / cm

Generally the slit width is mathematically represented as

              d =  \frac{1}{N  }

=>          d =  \frac{1}{ 5000  }

=>          d =   0.0002 \  c m  =  2.0 *10^{-6} \ m

Generally the condition for constructive interference for the first ray is mathematically represented as

         d sin(\theta_1) =  n *  \lambda_1

=>      \theta_1 = sin^{-1} [\frac{ 2 *  \lambda }{d}]

=>       \theta_1 = sin^{-1} [\frac{ 2 *   650 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_1 = 40.5 ^o

Generally the condition for constructive interference for the second ray is mathematically represented as

         d sin(\theta_2) =  n *  \lambda_2

=>      \theta_2 = sin^{-1} [\frac{ 2 *  \lambda_1 }{d}]

=>       \theta_2 = sin^{-1} [\frac{ 2 *   420 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_2 = 24.8  ^o

Generally the angular separation is mathematically represented as

            \theta  =  \theta_1 - \theta_1

=>          \theta  = 42.5^o -  24.8^o

=>          \theta  = 15.7^o

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