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Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails (1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt) . (a) Calculate the rate of temperature increase in degrees Celsius per second $\left(^{\circ} \mathrm{C} / \mathrm{s}\right)$ if the mass of the reactor core is $1.60 \times 10^{5} \mathrm{kg}$ and it has an average specific heat of 0.3349 $\mathrm{kJ} / \mathrm{kg}^{\circ} \cdot \mathrm{C}$. (b) How long would it take to obtain a temperature increase of $2000^{\circ} \mathrm{C}$ metals holding the radioactive materials to melt? (The initialrate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the $5 \times 10^{5}$ -kg steel containment vessel would also begin to heat up.)

(a) $2.80^{\circ} \mathrm{C} / \mathrm{s}$(b) 11.9 $\mathrm{m}$

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

Lectures

02:20

A solid, liquid, or gas is one of the three main states of matter. Solids have a definite shape and volume, and retain their shape when a force is applied to them. They are rigid, and do not flow to take on the shape of a container, but retain their own shape. Solids are held together by intermolecular forces, which are usually chemical bonds. Liquids have a free-flowing, continuous surface, and take the shape of a container. They flow to fill an available space. Their particles do not have a definite shape or volume, and they are not rigid. Gases have no definite shape or volume. They are not held together, and are not rigid. They flow to fill an available space. Gases are often described as being the state of matter with the lowest density.

06:13

Matter is a state of aggregation for a collection of atoms, molecules, ions, or particles. It is a general term for the substance of which all physical objects consist.

03:59

Even when shut down after …

04:29

03:00

02:51

09:26

A nuclear-fueled electric …

02:59

02:58

A typical nuclear reactor …

02:45

A nuclear power plant gene…

in this question. We want to calculate the temperature increase of a reactor core the rate of that in degree Celsius per second. So in this question, they give us the power output of the core when the cooling systems down. So you can start with that. We can write that as P, which is straight on here. Power is energy per time. So in our case, it's the heat output over specific period of time. They give that to us as 150 megawatts, which is the same as 150 times 10 to the six for Mega and Watts is just defined as Jules per second. Okay, so we're also giving the mass of the reactor core, which is on 0.6 times 10 to the fifth ah, kilograms. And we are also given the specific heat, See, which is your 0.33 for nine killa jewels for kilogrammes degrees Celsius. We're gonna want to use our standard heat equation. Q equals emcee Delta T. And we're gonna solve this for the he the change in temperature delta T per second. So, you know, cute per second. That's just the power output that we were given. So let's rearrange this. Divide the other side by EMC to get Delta T equals Q over M. C. Okay, so we're just gonna solve that over here. So, Delta T, this is per second. We need the q per second, which is P. So that's 150 times 10 of the six Jules per second. We're gonna divide this by m times C. So that's 1.6 times 10 to the five kilograms time. See, which is your 0.3349 killer. Jules. But we have a jewels on top. So let's just write that out. Kill Oh is tended the three, sometimes end of the three Jules per kilogrammes degrees Celsius. So are kilograms. Cancel out her jewels. Cancel out with the one on top. We're left with degree Celsius, which comes up top on a second. So if you saw for this, um, those numbers and your calculator work out to 2.8 degrees Celsius per second. So that's the answer to the first part of our problem. And then the second part wants to know how long this it would take for this heat output to give us a temperature increase of 2000 degrees Celsius so we can use the same equation. Um, in this case, we have Q equals M c. Adult T. Okay, so we want the amount of time we have the power rights. We want the power times time, which will give us Q cause and see a delta t will just repeat this. So in this case, T equals M c Delta T over Pete. So that is 1.6 times 10 to the five kilograms I'm see is 3 34.9 ones were taken to cut the 1000 jewels per kilogrammes, degrees Celsius. Delta T is 2000 degrees Celsius from the problem statement. And we're gonna divide all this by the power, which is 150 times 10 to the sixth Jules per second. So you have jewels that cancel out. If Celsius that cancels out kilograms, that cancels out. We're just left with a seconds in the numerator. Saul, For this we get 714 0.45 seconds, which would be our answer. If you wantto get minutes out of that for something more understandable, we can divide that by 60 since their 60 seconds in a minute, which is roughly 11.9 minutes

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