Una muestra de agua de 250g de agua se caliento desde 10°C hasta 60°C calcula la cantidad de calor absorbido por el agua el calor especifico del agua es de 4186J/Kj. K

Calor absorbido, Q = 52325 Joules

Explanation:

Dados los siguientes datos;

Masa = 250g to kg = 250/1000 = 0.25kg

Temperatura inicial, T1 = 10°C to Kelvin = 273 + 10 = 283K

Temperatura final, T2 = 60°C to Kelvin = 273 + 60 = 333K

Capacidad calorífica específica = 4186 j/kg.K.

Para encontrar la cantidad de calor necesaria;

La capacidad calorífica viene dada por la fórmula;

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 333 - 283

dt = 50 K

Sustituyendo en la fórmula, tenemos;

Calor absorbido, Q = 52325 Joules

the part of the wave he is travelling on is the crest

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(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.