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Stolb23 [73]
3 years ago
11

3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.

Physics
1 answer:
Scrat [10]3 years ago
3 0

Answer:

514 cal

Explanation:

In order to calculate the lost heat by the amount of water you first take into account the following formula:

Q=mc(T_2-T_1)         (1)

Q: heat lost by the amount of water = ?

m: mass of the water

c: specific heat of water = 1cal/g°C

T2: final temperature of water = 11°C

T1: initial temperature = 12°C

The amount of water is calculated by using the information about the density of water (1g/ml):

m=\rho V=(1g/ml)(514ml)=514g

Then, you replace the values of all parameters in the equation (1):

Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal

The amount of water losses a heat of 514 cal

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Which of the below is an example of mimicry that enables prey species avoid predation?
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A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

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Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

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Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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