   Chapter 11, Problem 69AP

Chapter
Section
Textbook Problem

A “solar cooker” consists of a curved reflecting mirror that focuses sunlight onto the object to be heated (Fig. P11.69). The solar power per unit area reaching the Earth at the location of a 0.50-m-diameter solar cooker is 600. W/m2. Assuming 50% of the incident energy is converted to thermal energy, how long would it take to boil away 1.0 L of water initially at 20.°C? (Neglect the specific heat of the container.) Figure P11.69

To determine
The time taken to boil away water.

Explanation

Section 1:

To determine : The mass of the water.

Answer: The mass of water is 1.0 kg

Explanation :

Given Info: Volume of water is 1.0 L.

Formula to calculate the mass of the water is,

m=ρV

• m is the mass of water,
• ρ is the density of water,
• V is the volume of water,

Substitute 1000kg/m3 for ρ , 1.0 L for V to find m.

m=(1000kg/m3)(1.0liter)(103m31liter)=1.0kg

The mass of water is 1.0kg .

Section 2:

To determine : The energy required to boil away the water.

Answer: The energy required to boil away the water is 2.59×106J .

Explanation :

Given Info: Initial temperature of water is 20.0°C and initial temperature of the water is 20°C .

First, the heat supplied must increase the temperature of water till the water reaches its boiling point and then the heat supplied must transform all the water to steam at its boiling point 100°C .

Formula to calculate the energy required to increase the temperate up to the boiling point is,

Qboil=(mwatercwater)(TfTi)+mwaterLv

• Qboil is the heat required to raise the temperature of water to 100°C
• mwater is the mass of water,
• cwater is the specific heat capacity of water,
• Ti is the initial temperature of water and aluminum pot,
• Tf is the boiling temperature of water,
• Lv is the latent heat of vaporization of water,

Substitute 1.0 kg for mwater , 4186J/kg°C for cwater , 2.26×106J/kg for Lv , 20°C for Ti , and 100°C for Tf to find Qboil .

Qboil=[(1.0kg)(4186J/kg°C)(100°C20°C)]+[(1.0kg)(2

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