## Problem 2### Problem Statement:97. As we drill down into the rocks of Earth's crust, the temperature typically increases by 3.0°C for every 100 m of depth. Oil wells are commonly drilled to depths of 1830 m. If water is pumped into the shaft of the well, it will be heated by the hot rock at the bottom and the resulting heated steam can be used as a heat engine. Assume that the surface temperature is 20°C.(a) Using such a 1830-m well as a heat engine, what is the maximum efficiency possible?(b) If a combination of such wells is to produce a 2.5-MW power plant, how much energy will it absorb from the interior of Earth each day?---### Explanation:This problem involves geothermal energy and heat engine efficiency. The temperature increase with depth and the use of steam generated from heated water in an oil well to understand energy conversion is explored. - **(a) Maximum Efficiency Calculation:** The maximum efficiency of a heat engine can be calculated using the Carnot efficiency formula: \[ \eta = 1 - \frac{T_C}{T_H} \] where: - $ T_C $ is the temperature of the cold reservoir (surface temperature), - $ T_H $ is the temperature of the hot reservoir (bottom of the well).- **(b) Energy Absorption Calculation:** The energy absorbed from the Earth (heat input) to sustain a 2.5-MW power plant, taking into account the efficiency calculated in part (a).### Additional Notes:- **Temperature Increase:** Depth of the well = 1830 m Temperature gradient = 3.0°C per 100 m Increase in temperature = $ 1830 \, \text{m} \times \frac{3.0°C}{100 \, \text{m}} $- **Surface Temperature = 20°C**These calculations will involve converting temperatures into Kelvin for the Carnot efficiency calculation.This problem introduces concepts of earth science related to geothermal gradients and practical applications in energy engineering.

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Answered: 97. As we drill down into the rocks of…

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