Question

\(1.5 \mathrm{kg}\) of liquid water initially at \(12^{\circ} \mathrm{C}\) is to be heated at \(95^{\circ} \mathrm{C}\) in a teapot equipped with a \(800-\mathrm{W}\) electric heating element inside. The specific heat of water can be taken to be \(4.18 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{C}\) and the heat loss from the water during heating can be neglected. The time it takes to heat water to the desired temperature is \((a) 5.9 \mathrm{min}\) (b) 7.3 min \((c) 10.8 \mathrm{min}\) \((d) 14.0 \mathrm{min}\) \((e) 17.0 \mathrm{min}\)

Short answer

(Neglect heat loss.) Answer: (c) 10.8 min

Step-by-step solution

Write down the energy conservation equation

We will apply the energy conservation principle, where the heat energy provided by the heating element is equal to the heat energy absorbed by the water. The formula to calculate the heat energy absorbed by the water is \(Q = mcΔT\), where \(Q\) is the heat energy, \(m\) is the mass of water, \(c\) is the specific heat of water, and \(ΔT\) is the temperature change.

Plug in the given values

We have the mass of water \(m = 1.5 \mathrm{kg}\), specific heat of water \(c = 4.18 \mathrm{kJ / kg \cdot ^{\circ}C}\), initial temperature \(T_{i} = 12^{\circ}\mathrm{C}\), and final temperature \(T_{f} = 95^{\circ}\mathrm{C}\). The power of the heating element is given as \(P = 800\,\mathrm{W}\). First, calculate the temperature change \(ΔT = T_{f} - T_{i} = 95^{\circ}\mathrm{C} - 12^{\circ}\mathrm{C} = 83^{\circ}\mathrm{C}\).

Calculate the heat energy required to heat the water

Using the formula \(Q = mcΔT\), we can calculate the heat energy required to heat the water: \(Q=1.5\,\mathrm{kg}\times4.18\,\mathrm{kJ/kg\cdot^\circ C}\times83\,^\circ\mathrm{C}=518.37\,\mathrm{kJ}\).

Convert the power to kilojoules per minute

The power of the heating element is given in watts. Convert this to kilojoules per minute: \(P = 800\,\mathrm{W}\times\frac{1\,\mathrm{kJ}}{1000\,\mathrm{W}}\times\frac{60\,\mathrm{s}}{1\,\mathrm{min}} = 48\,\mathrm{kJ/min}\).

Calculate the time required to heat the water

Now, we will use the energy conservation equation: \(Q = Pt\) where \(t\) is the time in minutes. Solve for \(t\): \(t = \frac{Q}{P} = \frac{518.37\,\mathrm{kJ}}{48\,\mathrm{kJ/min}}=10.8\,\mathrm{min}\). So, the time it takes to heat the water to the desired temperature is \((c)\,10.8\,\mathrm{min}\).