Chapter 6, Problem 143IP

A cubic piece of uranium metal (specific heat capacity = 0.117 J/°C • g) at 200.0°C is dropped into 1.00 L deuterium oxide ("heavy water;" specific heat capacity = 4.21 I J/°C • g) at 25.5°C. The final temperature of the uranium and deuterium oxide mixture is 28.5°C. Given the densities of uranium (19.05 g/cm³) and deuterium oxide (1.11 g/mL). what is the edge length of the cube of uranium?

Interpretation Introduction

Interpretation: The edge length of the cube of uranium should be determined.

Concept Introduction:

           The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

                                           $\text{C = }\frac{\text{Heat absorbed}}{\text{Temperature change}}\mathrm{......}\text{(1)}$

         Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$\text{Absorbed heat }\left(\text{J}\right)\text{=}\text{Specific heat capacity×Temperature change}\left(\text{c}\right)\text{×mass ofsubstance }\left(\text{g}\right)\mathrm{......}\text{(2)}$

For the above equation heat is:

                                    $\text{S = q×M×T}\mathrm{.....}\text{(1)}$

                                                      q is heat (J)

                                                      M is mass of sample $\left(\text{g}\right)$

                                                      S is specific heat capacity $\left(\text{J/°C·g}\text{.}\right)$

                                                      T is temperature change $\left(\text{C}\right)$

For the process no heat loss to the surroundings means then the heat is

                   $\left(\text{absorbed}\right)\text{-q×M×ΔT=-q×M×ΔT}\text{(released)}\mathrm{......}\text{(3)}$

Answer to Problem 143IP

Answer

                The edge length of the cube of uranium is $\text{3}\text{.3 cm}$.

Explanation of Solution

Explanation

Given data:

                 Specific heat capacity of uranium $\text{= 0}\text{.117}\text{ J/°C· g}$

                 Specific heat capacity of heavy water $\text{= 4}\text{.18 J/°C · g}$

                 Specific heat capacity of steam $\text{= 2}\text{.02J/°C· g}$

     Volume of heavy  water is $\text{1}\text{.0}\text{L}$

                 Initial temperature of uranium is $\text{200°C}$

                 Final temperature of the mixture is $\text{28}\text{.5°C}$.

                            $\begin{array}{l}\text{Density of uranium is 19}{\text{.05 g/cm}}^{\text{3}}\\ \text{Density}\text{of}\text{heavy}\text{water}\text{is}\text{1}\text{.11 g/mL}\end{array}$

To calculate the required mass of heavy water.

                      $\begin{array}{l}\text{Mass}\text{of}\text{heavy}\text{water}\text{=}\text{volume×density}\\ \text{=}\text{1}{\text{.00×10}}^{\text{3}}\text{mL×}\frac{\text{1}\text{.11g}}{\text{mL}}\\ \text{=}\text{1110}\text{g}\end{array}$

• The volume of heavy water is multiplied by the density to give the mass of heavy water.
• The mass of heavy water is $\text{1110}\text{g}$.

To calculate the heavy water gained heat.

                                            $\begin{array}{l}{\text{q}}_{\text{(gain)}}\text{=}\text{4}\text{.211}\frac{\text{J}}{\text{°C}\text{.g}}\text{×1110g×}\left(28.5-25.2\right)\text{°C}\\ \text{=}1.4{\text{×10}}^4\text{J}\end{array}$

• The calculated mass of heavy water, given specific heat capacity and temperature change are plugging in to equation (1) to give a heat gain by heavy water.
• The heat gain by heavy water is $1.4{\text{×10}}^4\text{J}$.

To calculate the mass of uranium.

                    $\left(\text{absorbed}\right){\text{-q}}_{\text{Heavy}\text{water}}{\text{×M×ΔT=-q}}_{\text{uranium}}\text{×M×ΔT}\text{(released)}$

$\begin{array}{l}\text{Heat}\text{loss}{}_{\text{uranium}}=\text{0}\text{.117}\frac{\text{J}}{\text{°C}\text{.g}}\text{×mass×}\left(\text{200-28}\text{.5}\right)\text{°C}\\ \text{ =1}{\text{.4×10}}^{\text{4}}\text{J}\\ {\text{Mass}}_{\text{uranium}}\text{=}\frac{\text{1}{\text{.4×10}}^{\text{4}}}{\text{0}\text{.117}\text{×}\text{171}\text{.5}}\\ \text{=}\text{7}{\text{.0×10}}^{\text{2}}\text{g}\end{array}$

                                                

• The calculated gained heat of heavy water, given specific heat capacity and temperature change are plugging in to equation (1) and do some rearrangement to give a mass of uranium.
• The mass of uranium is $\text{7}{\text{.0×10}}^{\text{2}}\text{g}$.

To calculate the edge length of the cube of uranium.

                   $\begin{array}{l}\text{Cube edge}\text{=}\text{Mass×Density}\\ \text{Cube edge}{}_{\text{uranium}}\text{=}\text{7}{\text{.0×10}}^{\text{2}}\text{g×}\frac{\text{1}{\text{cm}}^{\text{3}}}{\text{19}\text{.05}\text{g}}\\ \text{=}\text{37}{\text{cm}}^{\text{3}}\\ \text{=}{\text{(37}}^{}\\ \text{=}\text{3}\text{.3}\text{cm}\end{array}$

• The calculated mass and density are multiplied to give volume uranium cube then this do sum math's to give a edge length of the cube of uranium.
• The edge length of the cube of uranium is $\text{3}\text{.3}\text{cm}$.

Conclusion

Conclusion

The edge length of the uranium cube was calculated by using calculated mass of heavy water and uranium and the length of the cube of uranium was found to be $\text{3}\text{.3}\text{cm}$.