Question

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)

Answer 1

The increases temperature is 28.66°C.

Step-by-step explanation:

Given that,

Efficiency = 25.0%

Intensity = 1000 W/m^2

1 h = 3600 sec

Area
`A= 4.00\ m^2`

We need to calculate the power

Using formula of intensity

`I=(P)/(A)`

`P=I*A`

Where, P = power

I = intensity

A = area

Put the value into the formula

`P = 1000*4.00`

`P=4000\ W`

We need to calculate the temperature

Now, Using formula of power

`P=(W)/(t)`

`4000*0.25=(mc\Delta T)/(t)`

`\Delta T=(0.25*4000*3600)/(30.0*4186)`

`\Delta T=28.66^(\circ)\ C`

Hence, The increases temperature is 28.66°C.