Answer:
a. 0.599
Explanation:
[tex]k_{al}[/tex] = Thermal conductivity of aluminum = 238 J/(s·m·°C)
[tex]k_{cu}[/tex] = Thermal conductivity of copper = 397 J/(s·m·°C)
[tex]A[/tex] = Area of each rod
[tex]d[/tex] = thickness of each rod
[tex]\DeltaT[/tex] = Difference in temperature for each rod
[tex]t[/tex] = time interval of use for each rod
Heat transferred by aluminum rod is given as
[tex]Q_{al} = \frac{k_{al} A\Delta T t}{d}[/tex]
Heat transferred by copper rod is given as
[tex]Q_{cu} = \frac{k_{cu} A\Delta T t}{d}[/tex]
Ratio of the heat transferred is given as
[tex]Ratio = \frac{Q_{al}}{Q_{cu}} = \frac{\frac{k_{al} A\Delta T t}{d}}{\frac{k_{cu} A\Delta T t}{d}} = \frac{k_{al}}{k_{cu}} = \frac{238}{397} = 0.599[/tex]
