Question

Ten milligrams of pure polonium-210 is placed in 500 g of water.If no heat is allowed to escape to the surroundings, how much will the temperature rise in 1 hour?

Short answer

Change in temperature$\Delta T=2.47°C$

Step-by-step solution

Decay constant of polonium-210

Given,

• Mass of polonium-210$=10mg=10\times {10}^{-6}kg$
• Mass of water$m_w=500g=500\times {10}^{-3}kg$
• Half-life of polonium$t_{\frac{1}{2}}=140days=3360hours$
• Time period t = 1 hour

${}_{84}{}^{210}Po\to {}_2{}^{4}He+{}_{82}{}^{206}Pb$

Now, let us find out the initial number of polonium atoms localid="1658423854044" $N_0$that would have been at the start of the decay

$$\begin{gathered} N_0=\frac{m_{p0}}{Z_p} \\ =\frac{{\displaystyle 10\times {10}^{-6}kg}}{{\displaystyle 209.982848u\times 1.66\times {10}^{-27}kg/u}} \\ {N}_0=2.86885\times {10}^{19} \end{gathered}$$

Let us find the decay constant$\lambda$ of polonium-210:

$\begin{array}{rcl}\lambda & =& \frac{In2}{t_{\frac{1}{2}}}\\ & =& \frac{In2}{3360}\\ & =& 2.062\times {10}^{-4}\text{decays/year}\\ & & \end{array}$

Finding the number of polonium atoms left after 1 hour of decay.

As we know:

$N=N_0{e}^{-\lambda t}$

Substituting the values and we get,

$\begin{array}{rcl}N& =& 2.86885\times {10}^{19}\times e^{-2.062\times {10}^{-4}}\\ & =& 2.86825\times {10}^{19}\\ N& =& 8.583\times {10}^{19}\\ & & \end{array}$

We have to find out the amount of kinetic energy released in the decay:

$\begin{array}{rcl}Q& =& (m_i-m_f)c^2\\ & =& (209.982873673-205.97446527-4.002603254)931.5MeV\\ & =& 5.407MeV\\ & & \end{array}$

Let us find out the amount of polonium-210 nuclei decayed as:

$\begin{array}{rcl}{\text{N}}_d& =& N_0-N\\ & =& 0.60\times {10}^{14}\\ & & \end{array}$

One polonium-210 nuclei releases 5.407 MeV of energy, then $N_d$nuclei releases:

$\begin{array}{rcl}{Q}^{\text{'}}& =& N_d\times Q\\ & =& 0.060\times {10}^{14}\times 5.407MeV\\ & =& 3.244\times {10}^{14}MeV\\ & =& 5190J\\ & & \end{array}$

Change in temperature of the water

Specific heat capacity of water$c=4200J/kg·C$

$\begin{array}{rcl}\Delta T& =& \frac{Q^{\text{'}}}{Cm}\\ \Delta T& =& \frac{5190}{4200\times 0.5}\\ \Delta T& =& 2.47°C\\ & & \end{array}$

Therefore, change in temperature$\Delta T=2.47°C$