The United States generates about \(5.0 \times 10^{16} \mathrm{J}\) of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost \(100 \%\) efficiency by an electric motor. (a) If this energy is generated by power plants with an average efficiency of \(0.30,\) how much heat is dumped into the environment each day? (b) How much water would be required to absorb this heat if the water temperature is not to increase more than \(2.0^{\circ} \mathrm{C} ?\)

The efficiency formula is given by the following equation: Efficiency = (output energy) / (input energy) Given that the energy generated by the power plants is equivalent to work, we have output energy as \(5.0 \times 10^{16} \mathrm{J}\) and efficiency as \(0.30\). We can solve for the input energy: 0.30 = \((5.0 \times 10^{16} \mathrm{J})\) / (input energy) input energy = \((5.0 \times 10^{16} \mathrm{J}) / 0.30\) input energy = \(1.67 \times 10^{17} \mathrm{J}\)

We know that the heat dumped into the environment is the difference between the input energy and the output energy. Heat dumped = input energy - output energy Heat dumped = (\(1.67 \times 10^{17} \mathrm{J}\)) - (\(5.0 \times 10^{16} \mathrm{J}\)) Heat dumped = \(1.17 \times 10^{17} \mathrm{J}\)

Given that water should not increase in temperature by more than 2.0°C, we can use the specific heat formula: \(\Delta Q = mc\Delta T\) Where \(\Delta Q\) is the heat absorbed, \(m\) is the mass of the water, \(c\) is the specific heat of water, and \(\Delta T\) is the change in temperature. We are given that \(\Delta T = 2.0^{\circ} \mathrm{C}\). The specific heat capacity of water \(c\) = \(4.18 \times 10^3 \mathrm{J/kg^{\circ}C}\). From our previous calculations, we have \(\Delta Q = 1.17 \times 10^{17} \mathrm{J}\). Now, we will solve for the mass \(m\) of the water needed to absorb the heat. \(1.17 \times 10^{17} \mathrm{J}\) = \(m (4.18 \times 10^3 \mathrm{J/kg^{\circ}C}) (2.0^{\circ} \mathrm{C})\) \(m = \dfrac{(1.17 \times 10^{17} \mathrm{J})}{(4.18 \times 10^3 \mathrm{J/kg^{\circ}C}) (2.0^{\circ} \mathrm{C})}\) \(m = 1.40 \times 10^{13} \mathrm{kg}\) So, to absorb the heat generated by the power plants, \(1.40 \times 10^{13} \mathrm{kg}\) of water would be needed without increasing the water temperature by more than 2.0°C.