\(1.2 \mathrm{~kg}\) of liquid water initially at \(15^{\circ} \mathrm{C}\) is to be heated to \(95^{\circ} \mathrm{C}\) in a teapot equipped with a \(1200-\mathrm{W}\) electric heating element inside. The teapot is \(0.5 \mathrm{~kg}\) and has an average specific heat of \(0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Taking the specific heat of water to be \(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated.

First, we need to determine the heat energy absorbed by the water. We can do this using the formula: \(Q_{water} = m_{water} \times c_{water} \times (\Delta T_{water})\) Where: - \(Q_{water}\) is the heat energy absorbed by the water - \(m_{water}\) is the mass of the water (\(1.2 \mathrm{~kg}\)) - \(c_{water}\) is the specific heat capacity of water (\(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\)) - \(\Delta T_{water}\) is the temperature difference experienced by the water (80°C) We can now plug in the values and calculate the heat energy required to heat the water: \(Q_{water} = 1.2 \mathrm{~kg} \times 4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \times 80 \mathrm{~K} = 402.048 \mathrm{~kJ}\)

Similarly, we now need to determine the heat energy absorbed by the teapot using the formula: \(Q_{teapot} = m_{teapot} \times c_{teapot} \times (\Delta T_{teapot})\) Where: - \(Q_{teapot}\) is the heat energy absorbed by the teapot - \(m_{teapot}\) is the mass of the teapot (\(0.5 \mathrm{~kg}\)) - \(c_{teapot}\) is the specific heat capacity of the teapot (\(0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\)) - \(\Delta T_{teapot}\) is the temperature difference experienced by the teapot (80°C) We can now plug in the values and calculate the heat energy required to heat the teapot: \(Q_{teapot} = 0.5 \mathrm{~kg} \times 0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \times 80 \mathrm{~K} = 28 \mathrm{~kJ}\)

To find the total heat energy required to heat the water and the teapot, we simply add the heat energy required for each component: \(Q_{total} = Q_{water} + Q_{teapot} = 402.048 \mathrm{~kJ} + 28 \mathrm{~kJ} = 430.048 \mathrm{~kJ}\)

Finally, we can find the time it takes for the water to be heated by dividing the total heat energy required by the power of the heating element. Note that we also need to convert the power from watts to kilowatts by dividing it by 1000: \(P = 1200 \mathrm{~W} = 1.2 \mathrm{~kW}\) Time \(t\) is given by: \(t = \frac{Q_{total}}{P} = \frac{430.048 \mathrm{~kJ}}{1.2 \mathrm{~kW}} = 358.373 \mathrm{~s}\) The water and the teapot will be heated to the desired temperature in approximately 358.373 seconds (or about 6 minutes).