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
Here are the calculations for the specific heat capacities of Zinc and Brass. Zinc Step 1: State the principle of calorimetry. Heat loss by zinc = Heat gained by water m_Z c_Z T_Z = m_W c_W T_W Step 2: Rearrange the formula to solve for the specific heat capacity of zinc (c_Z). c_Z = (m_W c_W T_W)/(m_Z T_Z) Step 3: Substitute the given values into the formula. Given: m_W = 0.303 \, kg c_W = 4180 \, J/kg K T_W = 1 \, °C = 1 \, K m_Z = 0.08 \, kg T_Z = 73 \, K c_Z = (0.303 \, kg)(4180 \, J/kg K)(1 \, K)(0.08 \, kg)(73 \, K) Step 4: Calculate the specific heat capacity of zinc. c_Z = 1266.54 \, J5.84 \, kg K c_Z = 216.8732... \, J/kg K Rounding to two decimal places: c_Z = 216.87 J/kg K Brass Step 1: State the principle of calorimetry. Heat loss by brass = Heat gained by water m_B c_B T_B = m_W c_W T_W Step 2: Rearrange the formula to solve for the specific heat capacity of brass (c_B). c_B = (m_W c_W T_W)/(m_B T_B) Step 3: Substitute the given values into the formula. Given: m_W = 0.277 \, kg c_W = 4180 \, J/kg K T_W = 1.5 \, °C = 1.5 \, K m_B = 0.08 \, kg T_B = 72.5 \, K c_B = (0.277 \, kg)(4180 \, J/kg K)(1.5 \, K)(0.08 \, kg)(72.5 \, K) Step 4: Calculate the specific heat capacity of brass. c_B = 1737.69 \, J5.8 \, kg K c_B = 299.6017... \, J/kg K Rounding to two decimal places: c_B = 299.60 J/kg K