Calculations

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To determine whether or not the block would make a decent heater, we must calculate several properties of the system. These properties include the amount of energy required by the system over
its operational period, the energy output in watts of the prism into its surroundings, the mass and volume of paraffin needed to store the energy required to maintain the internal temperature of the system
over the operational period, and the heat output of the thermal block into the system in Watts.

First to determine the output from the prism to its surroundings, given that the area of the prism is 70.57m2 and the metric R value is 3.52m2K/W, and that the difference in temperature between the prism and its surroundings is 50K.
                                                                                                                              
                                                                                                                                Working equation                                                                                                                         

      P=A(Tprism - Tamb)/Rm


Solving
70.57m2 (300K-250K)/3.52m2K/W = 1002W



Now that the instantaneous heat output of the prism is found, the total energy lost to surroundings during the operational period, Q, can be found.

Working equation

Pt=Q

Solving
1002W(5400s) = 5.41*10^6 J


Now with the total energy known, we can solve for the mass of paraffin wax necessary to replace the lost energy over the operational period. With the specific heat of paraffin wax being 2.9 kJ/kg, and the
latent heat of fusion being 200kJ/ kg, the total mass , m, can be solved for.

Working equation

Cwax (m) (Thigh - Fwax) + Cwax (m) ( Fwax -Thigh) + Cfwax (m) = Q

Solving
2.9kJ/kg*(m)*30 + 2.9kJ/kg*(m)*10 + 200kJ/kg*m = 5.41*10^6J
m=17.12kg


Now to solve for the volume, V, given the density of .93g/cm

Equation

m/Dwax = V

Solving
17.12kg/.93g/cm3 = .0184 m


Assuming a block depth of .1m and a block width of .3m, the length of the block will be .613m in length. Since the length by width gives us a surface area of one side, the instantaneous heat output of the block can be calculated.

Working equation

P= A(Fwax - Tprism)/Rwax

Solving
(.1839m2)(310K-300K)/.4W/m2 K = 4.6W