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
Step 1: Calculate the heat supplied by the electric heater. The heat supplied by the heater is given by the formula Q = VIt. Given: Voltage V = 1.6\,V Current I = 0.1\,A Time t = 15\,min = 15 × 60\,s = 900\,s Q_heater = (1.6\,V)(0.1\,A)(900\,s) Q_heater = 144\,J Step 2: Calculate the heat gained by the ice. The problem states "a piece of ice of mass 2kg is added at a temperature of 2^" and also provides the "latent heat capacity of ice". Ice cannot exist at 2^. This is a common ambiguity in physics problems. We will assume the intention is that the ice is at 0^, melts, and then the resulting water warms up. If it were already water at 2^, the latent heat would not be relevant. First, calculate the heat required to melt the ice at 0^: Q_melt = m_ice L_f Given: Mass of ice m_ice = 2\,kg Latent heat of fusion of ice L_f = 336000\,Jkg^-1 Q_melt = (2\,kg)(336000\,Jkg^-1) Q_melt = 672000\,J Next, calculate the heat required to warm the melted ice (now water) from 0^ to the final temperature T_f: Q_warm, ice water = m_ice c_water (T_f - 0^) Given: Specific heat capacity of water c_water = 4200\,Jkg^-1K^-1 Q_warm, ice water = (2\,kg)(4200\,Jkg^-1K^-1)(T_f) Q_warm, ice water = 8400 T_f\,J The total heat gained by the ice is: Q_ice total = Q_melt + Q_warm, ice water Q_ice total = 672000 + 8400 T_f Step 3: Calculate the heat lost by the initial water and the copper calorimeter. The initial temperature of the water and the calorimeter is 30^. Heat lost by the initial water: Q_water loss = m_water c_water (T_initial, water - T_f) Given: Mass of initial water m_water = 40\,kg Initial temperature of water T_initial, water = 30^ Q_water loss = (40\,kg)(4200\,Jkg^-1K^-1)(30^ - T_f) Q_water loss = 168000(30 - T_f)\,J Heat lost by the copper calorimeter: Q_copper loss = m_copper c_copper (T_initial, copper - T_f) Given: Mass of copper calorimeter m_copper = 1.5\,kg Specific heat capacity of copper c_copper = 400\,Jkg^-1K^-1 Initial temperature of copper T_initial, copper = 30^ Q_copper loss = (1.5\,kg)(400\,Jkg^-1K^-1)(30^ - T_f) Q_copper loss = 600(30 - T_f)\,J The total heat lost by the warmer components is: Q_loss total = Q_water loss + Q_copper loss Q_loss total = 168000(30 - T_f) + 600(30 - T_f) Q_loss total = (168000 + 600)(30 - T_f) Q_loss total = 168600(30 - T_f) Q_loss total = 5058000 - 168600 T_f Step 4: Apply the principle of conservation of energy. Assuming negligible heat loss to the surroundings, the total heat supplied by the heater plus the heat lost by the warmer components equals the total heat gained by the colder components. Q_heater + Q_loss total = Q_ice total 144 + (5058000 - 168600 T_f) = 672000 + 8400 T_f 5058144 - 168600 T_f = 672000 + 8400 T_f Rearrange the equation to solve for T_f: 5058144 - 672000 = 8400 T_f + 168600 T_f 4386144 = 177000 T_f T_f = (4386144)/(177000) T_f ≈ 24.78047^ Rounding to two decimal places, the final temperature is: T_f ≈ 24.78^ The final temperature of the mixture is 24.78^. That's 2 down. 3 left today — send the next one.