"(II) An iron boiler of mass 180 kg contains 710 kg of water a...

Question

"(II) An iron boiler of mass 180 kg contains 710 kg of water at 18°C. A heater supplies energy at the rate of 58,000 kJ/h. How long does it take for the water

(a) to reach the boiling point, and "

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with heat transfer. In this problem, we have a 190 kg copper kettle which has a specific heat capacity of 390 joules per kilogram degrees Celsius. And it's initially filled with 720 kg of milk at 19 °C. If a burner provides energy at a rate of 59,000 kilojoules per hour, how long will it take the milk to reach its boiling point? That the specific heat capacity of milk is similar to that of water, which is 4186 joules per kilogram degrees Celsius. We're given four possible choices as our answers. Choice A is 1.1 hour. Choice B is 2.2 hours. Choice C is 3.2 hours and choice D is 4.2 hours. We're gonna begin by recalling how the heat being transferred to an object changes its temperature. So in that case, the heat is being absorbed by an object which is Q is going to be equal to the specific heat capacity C multiplied by the mass that object multiplied by the change in temperature of that object delta TFQ is equal to DM delta T, but we also have heat being added to the system. And so, in this case, um our keep being added to the system Q is going to be equal to, we're actually given the power or the rate at which energy is being transferred. So that's gonna be just equal to the power multiplied by the time. And so just like any sort of um energy power is energy per time. And so that means our heat here is just gonna be equal to our power multiplied by time. So what we can do is set, the heat is being transferred into the system equal to the heat that's being absorbed by the objects. So we're gonna have our heat being transferred to the system, which is the er P multiplied by T our time. And that's gonna be equal to are specific heat of the copper kettle which will label as piece of K and we'll multiply it by its mass which is MK and multiplied by our change in temperature or delta capital T that's gonna be heat absorbed by the kettle. But we also need to add to that. The heat is being absorbed by the milk. And so that's gonna be the specific heat capacity of the milk which is gonna be cm multiplied by the massive mill which is mm multiplied by our delta capital T, which is our change in temperature. Now, since both the kettle and the milk started off at 19 °C and it's going to go up to the boiling point or the uh milk. The delta Ts here are the same. So at this point, I just need to really solve for the time. So I'll have T is equal to and I can simplify this a little bit. So I'll have the DK multiplied by MK plus the CM multiplied by the MM. And that whole quantity is multiplied by delta T. And we'll divide that by the power P. So I have values for everything in that formula. So I can write those out all have our time. Little T is equal to for the specific heat of the specific heat capacity for the kettle. That's gonna be the 390 joules per kilogram degrees Celsius. I'll have the 390 J per kilogram degrees Celsius and that gets multiplied by the mass, which is the 190 kilograms. It will add to that the um specific heat capacity for the milk, which is that of water. And so that's gonna be the 4186 joules per kilogram. Three °C multiplied by the mass of the milk, which is the 720 pilgrims. And that all gets multiplied by our change in temperature. Well, the boiling point of milk is similar to that of water since it's mostly water. So it's gonna be 100 °C minus the starting temperature of 19 °C. And I divide the whole thing by the power that we were given. Now, we were given this in terms of kilojoules per hour. We'll actually want to have that in joules per hour. So that'll be 5.9 multiplied by 10 to the seven jewels per hour. So to convert kilojoules to jewels, you need to multiply by 1000. So now that we have those values in there, we can solve for our time. And this will have units of hours because we've left our in the denominator there as our units. And that's also what our voices were given in. And so I plug these values into my calculator. This turns out to be 4.2 hours. So this corresponds to answer D. So just to recap what we've done here, we've set the heat that's being transferred into the system equal to the heat that's being absorbed by the objects and use that to calculate the time that it takes for those objects to heat up to the final temperature. So I hope that this has been useful and I'll see you in the next video.

Struggling with Physics?

"(II) An iron boiler of mass 180 kg contains 710 kg of water at 18°C. A heater supplies energy at the rate of 58,000 kJ/h. How long does it take for the water

(a) to reach the boiling point, and "

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Calorimetry with Temperature and Phase Changes practice set

Struggling with Physics?

"(II) An iron boiler of mass 180 kg contains 710 kg of water at 18°C. A heater supplies energy at the rate of 58,000 kJ/h. How long does it take for the water(a) to reach the boiling point, and "

Watch next

Calorimetry with Temperature and Phase Changes practice set

"(II) An iron boiler of mass 180 kg contains 710 kg of water at 18°C. A heater supplies energy at the rate of 58,000 kJ/h. How long does it take for the water

(a) to reach the boiling point, and "