Problem $1$

The thermal diffusivity $()$ of a material is given by the expression below, where $V$ represents volume, $k$ is thermal conductivity, $m$ is mass and $c_p$ is specific heat.

$=\frac{v*k}{m*c_p},[\frac{m^2}{s}]$

For a given specimen, the volume was measured with a graduated cylinder providing a reading of $117$ ml $.$The resolution of the cylinder was $2$ ml $.$ The mass was measured with a scale $15$ times. The mean value obtained for these measurements was $0\mathrm{.}314$ kg and the standard deviation was $0\mathrm{.}0063$ kg $.$ The scale$-$presented an accuracy of $1\mathrm{.}25\%$ of the reading.

The specific heat for the material was reported as $890+-22\frac{J}{k}gC(95\%).$ The thermal conductivity is well known with a value of $235\frac{W}{m}C.$ Find the thermal diffusivity of the material and its uncertainty for a $95\%$ probability.

Solution $=9\mathrm{.}8386*{10}^{-5}\frac{m^2}{s}$

$U_{}=5\mathrm{.}086*{10}^{-6}\frac{m^2}{s}$