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3 years ago

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Se coloca agua en un recipiente de aluminio y se pone a calentar en una estufa que le suministra 230 kj, lo cual hace que la tem

peratura aumente a 16°C. Si en las mismas condiciones a la misma cantidad de agua se le suministra el doble de energía ¿cual sería el aumento de su temperatura?

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

Answer:

32 °C.

Explanation:

Hola.

En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:

$Q=mCp\Delta T$

De este modo, dado que estamos estudiando la misma sustancia (agua) con masa constante, la relación calor-temperatura es lineal y directamente proporcional, por tal razón, si se duplica el calor suministrado, la temperatura también será duplicada, de modo que:

$\Delta T_{nuevo}=2*16\°C\\\\\Delta T_{nuevo}=32\°C$

¡Saludos!

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Answer:

$a=2.4\ m/s^2$

Explanation:

Given that,

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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

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1 year ago

How is the Earth's magnetic field like a bar magnet's magnetic field?

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3 years ago

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

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Answer:

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Explanation:

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When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

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If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

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$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

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