Question

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150MW by the radio active decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails. (1 watt = 1 joule or 1W = 1J/s and 1MW = 1megawatt)Calculate the rate of temperature increase in degrees Celsius per second (°C/s) if the mass of the reactor core is 1.60×105kg and it has an average specific heat of 0.3349 kJ/kg°C.

Answer 1

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$