Measuring Radiation Dosage — Question 10

Question 10

Calorimetry measures dose through a very small temperature rise

An insulated absorber has specific heat $c=4000\,\mathrm{J/(kg\,K)}$ and undergoes radiation-induced temperature rise $\Delta T=.250\,\mu\mathrm K$. Assume all deposited energy becomes heat, with no heat loss. Independent relative standard uncertainties are 1.00 percent in c and 4.00 percent in temperature rise.

1. Derive and calculate absorbed dose without needing the absorber mass.

2. Find the relative and absolute standard uncertainties in dose.

Question 10 - Solution

Calorimetry measures dose through a very small temperature rise

(a) Method and calculation

$$E=mc\Delta T,\quad D=E/m\Rightarrow\boxed{D=c\Delta T}.$$$$\boxed{D=4000(.250\times10^{-6})=.00100\,\mathrm{Gy}=1.00\,\mathrm{mGy}}.$$

(b) Interpretation and further calculation

$$\boxed{\frac{\sigma_D}{D}=\sqrt{(.0100)^2+(.0400)^2}=.04123}.$$$$\boxed{\sigma_D=.04123\,\mathrm{mGy}}.$$

Physical check

The mass cancels because heat capacity and absorbed dose use the same material mass. Heat losses or chemical energy storage would require additional corrections.