To calculate the quantity of heat transferred to the rod, we can use the formula:
\[ Q = \text{Mass of coal burned} \times \text{Heat value of coal} \times \text{Percentage of heat transferred} \]
However, we need the heat value of the coal to conduct this calculation. The heat value can vary depending on the type of coal, but a common approximate value for sub-bituminous coal is about 24 MJ/kg.
First, let's convert the mass of coal burned from grams to kilograms:
\[ \text{Mass of coal burned} = 750 , \text{g} = 0.750 , \text{kg} \]
Now, we can calculate the total heat produced by the burned coal:
\[ Q_{\text{total}} = \text{Mass of coal burned} \times \text{Heat value of coal} \] \[ Q_{\text{total}} = 0.750 , \text{kg} \times 24 , \text{MJ/kg} = 18 , \text{MJ} = 18,000 , \text{kJ} \]
Now, we need to calculate the quantity of heat that is transferred to the rod, taking into account the 60% efficiency:
\[ Q_{\text{transferred}} = Q_{\text{total}} \times \text{Percentage of heat transferred} \] \[ Q_{\text{transferred}} = 18,000 , \text{kJ} \times 0.60 = 10,800 , \text{kJ} \]
Thus, the quantity of heat transferred to the rod is 10,800 kJ.