How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water initially at $11^\circ \textrm{C}$? Assume that the part of the pot which is heated with the water is made of 280 g of aluminum, and that no water boils away.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Power is energy absorbed per time. And we can rearrange this for time by multiplying both sides by t over p and the T's cancel on the right hand side and leaving us with Q over p there. And on the left hand side the p's cancel leaving us with t. So, time is the heat gained divided by the power. So, the heat gained is the heat gained by the water plus the heat gained by the pot. And they both have the same change in temperature because they have the same final temperature and they start with the same initial temperature. So, we can factor out this Δt. So is Δt times mass water times specific heat capacity of water plus the mass of the pot times the specific heat capacity of the pot. So, it ends at a temperature of 100 degrees Celsius and minus the initial temperature of 11 degrees Celsius times 0.75 liters of water times 1 kilogram per liter to get the mass of water times 4,186 joules per kilogram Celsius degree, specific heat capacity of water, times 0.28 kilograms of aluminum for the pot times 900 joules per kilogram Celsius degree, specific heat capacity of aluminum. And this gives 301,843 joules absorbed. And substitute that in for Q and divide by the power of the pot, 750 watts. And that gives 402.458 seconds times 1 minute for every 60 seconds gives us about 6.7 minutes for the pot to boil.