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

The temperature of the core raises by
`2.8^(o)C` every second.

Step-by-step 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* 10^(5)kg` will require

`0.3349* 1.60* 10^(5)kJ\\\\=0.53584* 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

`(150* 10^(6))/(0.53584* 10^(8))=2.8^(o)C/s`