is this correct for question 4 - Let's review the calculation again: Given data: Resistance of the heating element (Rheater) = 27.1 Initial temperature of water (Ti) = 22.2C Time (t) = 170 s Assuming a standard mains voltage of 240 V Using Equation (4): = 2 E= R V 2 t = ( 240 ) 2 170 27.1 E= 27.1 (240) 2 170 3589.47 E3589.47J Now, using Equation (1): = q=mCpT Given: Density of water at room temperature = 1.0 g/mL Mass of 1 L of water = 1000 g Final temperature (Tf) of the water is the boiling point, approximately 100C. = T=TfTi = 100 22.2 = 77.8 T=10022.2=77.8C Now, we can calculate the specific heat capacity (Cp) of water: = Cp= mT E = 3589.47 1000 77.8 Cp= 1000g77.8C 3589.47J 0.046 / Cp0.046J/gC