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A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2. What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2-area collector? (cw = 4 186 J/kg×°C)

Respuesta :

Answer:

The increases temperature is 28.66°C.

Explanation:

Given that,

Efficiency = 25.0%

Intensity = 1000 W/m^2

1 h = 3600 sec

Area [tex]A= 4.00\ m^2[/tex]

We need to calculate the power

Using formula of intensity

[tex]I=\dfrac{P}{A}[/tex]

[tex]P=I\timesA[/tex]

Where, P = power

I = intensity

A = area

Put the value into the formula

[tex]P = 1000\times4.00[/tex]

[tex]P=4000\ W[/tex]

We need to calculate the temperature

Now, Using formula of power

[tex]P=\dfrac{W}{t}[/tex]

[tex]4000\times0.25=\dfrac{mc\Delta T}{t}[/tex]

[tex]\Delta T=\dfrac{0.25\times4000\times3600}{30.0\times4186}[/tex]

[tex]\Delta T=28.66^{\circ}\ C[/tex]

Hence, The increases temperature is 28.66°C.

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