Spent steam from an electric generating plant leaves the turbines at 120.0°C and is cooled to 90.0°C liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling tower water for each kg of spent steam?I'm not sure how to do this one please help!!
To solve this problem, we use two simple equations concerning heat: Q=mc?T expresses the amount of heat needed to change an object from one temperature to another, where Q is the amount of heat, m is the mass of the material, c is the specific heat of the material, and ?T is the change in temperature (measured in °C or K--see comment below). You'll use this equation first for steam as it is cooled from 120 °C to 100 °C, then for water as it is cooled from 100 °C to 90 °C. However, at 100 °C, both steam and water exist. Here the water is vaporized to steam, or vice versa. We express the heat needed to convert water to steam: Q = m*L, where m is the mass of water converted to steam, and L is the latent heat of vaporization for water/steam. In this case, however, Q will be negative, as heat will be released in the cooling from steam to water. Sum all of these heats together and you'll get the total heat change of the system. This should be a negative number, as heat is released in all three heat calculations. Figures: Specific heat of water: 4186 [J/(kg °C)] or [J/(kg K)] Specific heat of steam: 2010 [J/(kg °C)] or [J/(kg K)] Latent heat of vaporization for water/steam: 2.26 x 10^6 [J/kg] The change in temperature can actually be expressed in either °C or K. Since Celsius and Kelvin use the same scale and differ only by a constant, and since we're taking the difference when we calculate ?T, it doesn't matter which of the two temperature scales you choose.
