A 800 g block of metal at 100 °C is dropped into 1.5 kg of water at 15 °C. The final temperature of the mixture is 20 °C. What is the specific heat capacity of the metal?

Here's the solution to question 11: Step 1: Identify the given values and the unknown. Given: • Mass of metal, m_m = 800 \, g = 0.8 \, kg • Initial temperature of metal, T_m,i = 100 \, °C • Mass of water, m_w = 1.5 \, kg • Initial temperature of water, T_w,i = 15 \, °C • Final temperature of the mixture, T_f = 20 \, °C • Specific heat capacity of water, c_w = 4200 \, J kg^-1 \, °C^-1 (standard value) Unknown: • Specific heat capacity of metal, c_m Step 2: Calculate the change in temperature for both the metal and the water. Change in temperature of metal: T_m = T_m,i - T_f = 100 \, °C - 20 \, °C = 80 \, °C Change in temperature of water: T_w = T_f - T_w,i = 20 \, °C - 15 \, °C = 5 \, °C Step 3: Apply the principle of calorimetry. According to the principle of calorimetry, the heat lost by the hot body (metal) is equal to the heat gained by the cold body (water), assuming no heat loss to the surroundings. Heat lost by metal = Heat gained by water Q_m = Q_w The formula for heat transfer is Q = mc T. m_m c_m T_m = m_w c_w T_w Step 4: Substitute the values into the equation and solve for c_m. (0.8 \, kg) × c_m × (80 \, °C) = (1.5 \, kg) × (4200 \, J kg^-1 \, °C^-1) × (5 \, °C) 64 \, c_m = 31500 c_m = (31500)/(64) c_m = 492.1875 \, J kg^-1 \, °C^-1 The specific heat capacity of the metal is 492.19 J kg^-1 °C^-1.