Even when shut down after a period of normal use, a large...

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 initial rate 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.)

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 initial
rate 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.)

Key Concepts

Power and Energy Conversion

Power is the rate at which energy is transferred or converted, measured in watts (W). One watt is equivalent to one joule per second (1 W = 1 J/s). This concept connects energy, measured in joules, with time, allowing the determination of how much energy is delivered over a given period. It is fundamental in calculating thermal energy transfer in systems where heat is produced or dissipated.

Specific Heat Capacity

Specific heat capacity is a material property that indicates the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It is used to relate the energy input (or extraction) from a material to its resulting temperature change, making it essential for understanding and calculating thermal energy storage and transfer in a substance.

Heat Transfer and Energy Balance Equation

The heat transfer process in a system can be analyzed using the energy balance principle, which is often expressed by the equation Q = mc?T, where Q is the heat energy, m is mass, c is the specific heat capacity, and ?T is the temperature change. This relation is crucial in determining the temperature change of a system when a known amount of energy is added or removed.

Rate of Temperature Change

The rate of temperature change in a system undergoing heating or cooling can be derived by combining the concept of power with the energy balance equation. By considering the power input and the material’s specific heat capacity and mass, one can calculate how quickly the temperature of a body will change over time, which is important for understanding the dynamics of thermal systems.

Time Calculation for Temperature Change

This concept involves determining the duration required to achieve a specified temperature change given a constant rate of energy transfer. Using the relationship between energy, mass, specific heat capacity, and temperature change, one can estimate the time needed for a system to reach a particular temperature increment, ensuring proper design and safety measures in thermal management applications.

Transcript

00:02 Okay, in this question, we want to calculate the temperature increase of a reactor core, the rate of that in degree celsius per second.

00:12 So in this question, they give us the power output of the core when the cooling system is down.

00:18 So we can start with that, and we can write that as p, which is straight down here, power is energy per time.

00:26 So in our case, it's the heat output over a specific period of time.

00:30 And they give that to us as 150 megawatts, which is the same as 150 times 10 to the 6 for mega.

00:42 And watts is just defined as joules per second.

00:48 Okay, so we're also giving the mass of the reactor core, which is 1 .6 times 10 to the fifth kilograms.

00:58 Kilograms and we are also given the specific heat c which is 0 .3349 kilojoules per kilograms degrees celsius.

01:15 We're going to want to use our standard heat equation q equals mc delta t and we're going to solve this for the heat the change in temperature delta t per second so we know q per second that's just the power output that we were given so let's rearrange this divide the other side by mc to get delta t equals q over mc okay so we are just going to solve that over here so delta t or this is per second we need the q per second which is p so that's 150 times 10 to the 6 joules per second we're going to divide this by m times c so that's 1 .6 times 10 to the 5 kilograms times c which is 0 .3349 kilojoules but we have a joules on top so let's just write that out.

02:19 Kilo is 10 to the 3 so times 10 to the 3 joules per kilograms degrees celsius so our kilograms cancel out our joules cancel out with the one on top we are left with the degree celsius which comes up top in a second.

02:36 So if you solve for this, those numbers in your calculator work out to 2 .8 degrees celsius per second.

02:45 So that's the answer to the first part of our problem...

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