This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Hey, good to see you again. Here's the solution to question 4. Step 1: Identify the given values. Mass of aluminum calorimeter, m_cal = 500 g = 0.5 kg Specific heat capacity of aluminum, c_cal = 900 J/kg K Mass of liquid X, m_X = 300 g = 0.3 kg Current, I = 3.0 A Voltage, V = 220 V Time, t = 2 minutes = 2 × 60 s = 120 s Initial temperature of liquid X and calorimeter, T_initial Final temperature of liquid X and calorimeter, T_final = 80^ C Step 2: Calculate the heat supplied by the electric heater. The heat supplied by the heater is given by the formula Q = VIt. Q = (220 V)(3.0 A)(120 s) Q = 79200 J Step 3: Calculate the heat gained by the aluminum calorimeter. The heat gained by the calorimeter is given by Q_cal = m_cal c_cal T. The temperature rise for the calorimeter is T = T_final - T_initial = (80 - T_initial)^ C. Q_cal = (0.5 kg)(900 J/kg K)(80 - T_initial) Q_cal = 450(80 - T_initial) J Step 4: Calculate the heat gained by liquid X. The heat gained by liquid X is given by Q_X = m_X c_X T. The specific heat capacity of liquid X is c_X = 2800 J/kg K. The temperature rise for liquid X is T = T_final - T_initial = (80 - T_initial)^ C. Q_X = (0.3 kg)(2800 J/kg K)(80 - T_initial) Q_X = 840(80 - T_initial) J Step 5: Apply the principle of conservation of energy. The heat supplied by the heater is equal to the total heat gained by the calorimeter and liquid X. Q = Q_cal + Q_X 79200 = 450(80 - T_initial) + 840(80 - T_initial) 79200 = (450 + 840)(80 - T_initial) 79200 = 1290(80 - T_initial) Step 6: Solve for the initial temperature, T_initial. 80 - T_initial = (79200)/(1290) 80 - T_initial ≈ 61.395 T_initial = 80 - 61.395 T_initial ≈ 18.605^ C Rounding to one decimal place: T_initial ≈ 18.6^ C The initial temperature of the liquid is 18.6^ C.