Question

1.2of liquid water initially at \(15^{\circ} \mathrm{C}\) is to be heated to \(95^{\circ} \mathrm{C}\) in a teapot equipped with a 1200 -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.

Short answer

Answer: It takes approximately 357.73 seconds (or about 6 minutes) to heat the water to 95°C.

Step-by-step solution

Determine the initial and final temperatures

The initial temperature of the water (\(T_{initial}\)) is \(15^{\circ} \mathrm{C}\). The final temperature (\(T_{final}\)) we want to achieve is \(95^{\circ} \mathrm{C}\).

Calculate the temperature difference

Calculate the difference between the initial and final temperatures: \(\Delta T = T_{final} - T_{initial}\) \(\Delta T = 95^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C}\) \(\Delta T = 80 \mathrm{K}\)

Determine the mass of water

The volume of liquid water is given as 1.2 L. Since the density of water is about \(1000 \mathrm{~kg/m^3}\), we can calculate the mass of water: \(mass_{water} = 1.2 \mathrm{~L} * \frac{1000 \mathrm{~kg}}{1000 \mathrm{~L}} = 1.2 \mathrm{~kg}\)

Calculate the energy required to heat the water

To calculate the amount of energy needed to heat the water, use the equation: \(Q_{water} = mass_{water} * c_{water} * \Delta T\) where \(c_{water} = 4.18 \mathrm{~kJ/kg \cdot K}\) is the specific heat of water. \(Q_{water} = 1.2 \mathrm{~kg} * 4.18 \mathrm{~kJ/kg \cdot K} * 80 \mathrm{~K} = 401.28 \mathrm{~kJ}\)

Calculate the energy required to heat the teapot

Similarly, calculate the amount of energy needed to heat the teapot: \(Q_{teapot} = mass_{teapot} * c_{teapot} * \Delta T\) where \(mass_{teapot} = 0.5 \mathrm{~kg}\), and \(c_{teapot} = 0.7 \mathrm{~kJ/kg \cdot K}\). \(Q_{teapot} = 0.5 \mathrm{~kg} * 0.7 \mathrm{~kJ/kg \cdot K} * 80 \mathrm{~K} = 28 \mathrm{~kJ}\)

Calculate the total energy required

Add the energy required to heat the water and the teapot: \(Q_{total} = Q_{water} + Q_{teapot}\) \(Q_{total} = 401.28 \mathrm{~kJ} + 28 \mathrm{~kJ} = 429.28 \mathrm{~kJ}\)

Determine the time required to heat the water

Lastly, calculate the time required by dividing the total energy required by the power of the heating element (1200-W or equivalently 1.2 kW): \(time = \frac{Q_{total}}{Power}\) \(time = \frac{429.28 \mathrm{~kJ}}{1.2 \mathrm{~kW}} = 357.73 \mathrm{~s}\) Therefore, it will take about 357.73 seconds (or approximately 6 minutes) to heat the water to \(95^{\circ} \mathrm{C}\).