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Tresset [83]
3 years ago
14

Calculate the change in thermal energy of the water in a pond with a mass of 1,000 Kg and a specific heat of 4,184 J/kg °C if th

e water cools by 1°C
​
Physics
2 answers:
bearhunter [10]3 years ago
7 0

Answer:

<em>4</em><em>1</em><em>8</em><em>4</em><em>0</em><em>0</em><em>0</em><em> </em><em>J</em>

Explanation:

that is the aNSweR

Inessa [10]3 years ago
7 0

We will consider that specific heat of any material raises the temperature of 1 kg in 1°C.

Q= <em>m</em> ( T<em>f</em><em> </em>- T<em>i</em> ) C,

where <em>m</em> is mass, T<em>f</em> is final temperature, T<em>i</em> is initial temperature and C is the specific heat of the substance.

So, by substituting values, the answer we get is <u>418400</u><u>0</u><u> J</u>

Hope it helps you...

Answered by Benjemin

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Algunas fabricas de balones de fútbol ubicadas en la costa inflan los balones que van a ser vendiéndose las ciudades como pasto,
otez555 [7]

Answer:

balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure

Explanation:

Este ejercicio es referente a la mecánica de fluidos, usemos la expresión para la presión  

       P = ρ g h

En es el caso del balón  usemos la presión en la pared extrema, llamemos P la presión por el gas en el interior y P_ext la presión atmosférica del lugar

        cuando se llena el valor en una ciudad de baja altura la presión atmosférica es mas alta

          P_int1 < P_ext1

por lo cual la pared del balón no se mantiene rígida.

Cuando el balón es trasladado a una ciudad con mayor altura sobre el nivel del mar la presión exterior disminuye

       P_ext2 = ρ g h₂ < P_ext1

en promedio la presión disminuye con la altura  en 0,029 atm cada 250 m

por lo tanto como la cantidad de gas es constante en el interior la presión interior del globo es constante y en esta ciudades se hace igual o un poco mayor que la presión atmosférica, en consecuencia la pared del globo esta rígida

        P_int2 >P_ext2

Traslate

This exercise is related to fluid mechanics, let's use the expression for pressure

       P = ρ g h

In the case of the balloon, let's use the pressure on the extreme wall, let's call P the pressure for the gas inside and P_ext the atmospheric pressure of the place

        when the value is filled in a low-lying city the atmospheric pressure is higher

          P_int1 <P_ext1

therefore the wall of the ball does not remain rigid.

When the ball is transferred to a city with higher altitude above sea level, the external pressure decreases

       P_ext2 = ρ g h <P_ext1

on average the pressure decreases with height by 0.029 atm every 250 m

therefore as balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure, therefore the wall of the

        Pint 2> Pe

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Explanation:

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