One example of a type of solar thermal system is solar cooking, where we use sunlight to heat food and water. With solar cooking, we're able to prepare food without using electricity or burning fuel! We can get the energy we need directly from sunlight.

Suppose we want to boil a kettle of water of volume \(V=\SI{1}{\liter},\) using sunlight of intensity \(I_\text{Sun} = \SI[per-mode=symbol]{1000}{\watt\per\meter\squared}\) from a concentrating mirror of area \(A=\SI{1}{\meter\squared}.\) How long will it take to heat that water from \(\SI{20}{\celsius}\) to \(\SI{100}{\celsius}\)?

**Assumptions and Details**:

- Ignore the thermal mass of the kettle (only consider the water) and ignore all heat losses from the kettle
- Assume all sunlight incident on the mirror ends up absorbed by the kettle
- The density and heat capacity of water are \(\rho_\text{water} = \SI[per-mode=symbol]{1}{\kilo\gram\per\liter}\) and \(C_\text{water} = \SI[per-mode=symbol]{4.18}{\joule\per\gram\per\kelvin},\) respectively.

In the last quiz, we learned that when an object absorbs thermal energy, its temperature increases. If you put an absorber in sunlight, it continuously absorbs more energy. Despite constantly absorbing energy, the temperature doesn't increase indefinitely.

The reason that the temperature of the object doesn't increase to infinity is because as the absorber's temperature increases, it begins to lose heat to the ambient environment, which is at a lower temperature. The absorber stops heating up once it reaches a temperature where the absorbed sunlight is balanced by losses to the surroundings.

One mechanism for an absorber to lose heat to its surroundings is through **convection**. Convective heat loss occurs from a surface that's exposed to a fluid at lower temperature. As the fluid near the surface heats up, it carries thermal energy away from the surface and is replaced by fluid at the ambient temperature.

The effectiveness of a fluid at pulling heat away from a surface through this process is characterized by a convection coefficient, measured in \(\si[per-mode=symbol]{\watt\per\meter\squared\per\kelvin}\). The convection losses are proportional to the temperature difference between the surface and the fluid.

In the case of a solar absorber surface, which geometric parameter do convection losses depend on?

The overall system efficiency for a solar thermal plant is given by electrical power output divided by the incident solar power. The solar receiver absorbs sunlight, and delivers some of that absorbed energy to the heat engine. The delivered heat is equal to the absorbed sunlight minus any thermal losses. The heat engine then converts some of the delivered heat to electricity, based on its efficiency.

In the previous quiz, we found that a solar receiver under a solar intensity of \(\SI[per-mode=symbol]{50000}{\watt\per\meter\squared}\) has a maximum operating temperature of \(\SI{971}{\kelvin}\). We calculated the Carnot efficiency associated with an operating temperature \(T_H = \SI{971}{\kelvin}\) to be \(69.1\%\). If we had a solar thermal system under a solar intensity of \(\SI[per-mode=symbol]{50000}{\watt\per\meter\squared}\) and operated it at a temperature of \(\SI{971}{\kelvin}\), what would the overall system efficiency \(\eta_{\textrm{sys}}\) be?

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