The burner on an electric stove has a power output of 2.0 kW. A 7...

Question

The burner on an electric stove has a power output of 2.0 kW. A 750 g stainless steel teakettle is filled with 20°C water and placed on the already hot burner. If it takes 3.0 min for the water to reach a boil, what volume of water, in cm^3 , was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer. Let's see what it's asking us. A three kilowatt electric heater is used to heat a copper pot filled with water that is initially at 30 degrees Celsius. It takes four minutes for the water to reach the boiling point. If the mass of the copper pot is one kg, determine the volume of water in centimeters cube that was present in that pot. Our multiple choice answers here in units of cubic centimeters are a 2.36 times 10 to the third B 3. times 10 to the third C 2.41 times 10 to the negative two or D 3.34 times 10 to the negative two. OK. So the first thing we're going to do is look at our heat energy balance. So the heat energy of the heater is going to equal that of the copper pot or Q sub C plus that of the water. Because we are in a closed system, we can recall that our equation for Q is MC delta T. And so this simplifies to uh we can also recall that for the heater in terms of power because we were given the power of the heater, that relationship is the power of the heater multiplied by time is that energy. And so this equation becomes power of the heater multiplied by time is equal to MC M sub C for the copper pot, C sub C delta T sub C plus. Sorry, let's get that a little bit neater there. So M sub C C sub C delta T of the copper pot plus the mass of the water, the specific heat capacity, sea of the water times delta T of the water. We are asked for the volume of the water. That's not a term that we see here, but the mass of the water is and from the mass, we can find the volume. So that is the term that we're going to be targeting. Um And, and solving for. So let's isolate that term. On the left hand side of the equation, we have mass of the water is equal to the power of the heater multiplied by the time of the heater minus the mass of the pot, the specific heat capacity of the pot and the delta T of the pot divided by the specific heat capacity of the water and the delta T of the water. And so one thing that we can note here is that the copper pot and the water are being heated at the same temperature. They're in the same, you know, they're, they're in the same vessel. So they have the same initial and final temperatures. So we can expand that term to simply be T F minus T initial. And that's going to be the same for both the pot and the water. So now it is time to plug in those values that we know for each of these terms. So the power of the heater was given as three kilowatts, we're going to keep that in standard units. So that becomes three times 10 to the third. And then we can recall that a watt is simply a duel per second multiplied by the time. So we're dealing with four minutes again, standard units, we will multiply that by 60 seconds per minute minus the mass of the copper pot was given in the problem as one kg. The specific heat capacity of copper is 385 joules per kilogram, Celsius and the delta T is 70 degrees Celsius. So our initial temperature was 30 degrees. So we had sorry, we have a, an initial temperature of 30 degrees Celsius, a final temperature of the boiling point of water, which we can recall is 100 degrees Celsius. And so that's how we come up with that 70 degrees Celsius as our delta T. OK. Now, all of that divided by the specific heat capacity of water, which we can recall is 4000, 190 joules per kilogram seed again multiplied by that delta T or 70 degrees C, we plug that into our calculator and we get a mass of the water of 200 or sorry 2.36 kg. And the last step is to recall that volume is related to mass by the equation, volume equals mass divided by density. And so the mass of the water we just sold for 2.36 kg multi or sorry, divided by the density kg per meter cubed for water. And so we come up with a volume of 2.36 times 10 to the third centimeters cubed. And that aligns with answer choice. A oops, I like that answer. Choice. A is the right answer for this problem. That's all we have for this one. We'll see you in the next video.

Struggling with Physics?

Watch next

Specific Heat & Temperature Changes practice set