To solve the problem, we need to calculate the rise in temperature of a substance when it falls a vertical distance and all its potential energy is converted into heat energy upon impact. Let's break this down step-by-step. ### Step 1: Calculate the Potential Energy (PE) When the substance falls from a height, it possesses gravitational potential energy, which can be calculated using the formula: \[ PE = mgh \] Where: - \( m \) = mass of the substance (in kg) - \( g \) = acceleration due to gravity (in m/s²) - \( h \) = height fallen (in m) In this case, we are not given the mass \( m \) of the substance, but we will see that it cancels out later in our calculations. Given: - \( g = 10 \, \text{m/s}^2 \) - \( h = 20 \, \text{m} \) Substituting the values into the formula, we get: \[ PE = m \cdot 10 \cdot 20 = 200m \, \text{J} \] ### Step 2: Relate Potential Energy to Heat Energy When the substance hits the ground, all of its potential energy is converted into heat energy. The heat energy (\( Q \)) gained by the substance can be expressed using the formula: \[ Q = mc\Delta T \] Where: - \( c \) = specific heat capacity of the substance (in J/kg·K) - \( \Delta T \) = rise in temperature (in °C or K) Given: - \( c = 450 \, \text{J/kg·K} \) ### Step 3: Set the Two Energies Equal Since all the potential energy is converted into heat energy, we can set the two equations equal to each other: \[ 200m = mc\Delta T \] ### Step 4: Cancel Out the Mass Since \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)): \[ 200 = c\Delta T \] ### Step 5: Solve for the Rise in Temperature (\( \Delta T \)) Now we can solve for \( \Delta T \): \[ \Delta T = \frac{200}{c} \] Substituting the value of \( c \): \[ \Delta T = \frac{200}{450} \] ### Step 6: Simplify the Calculation Now, simplifying \( \frac{200}{450} \): \[ \Delta T = \frac{200 \div 50}{450 \div 50} = \frac{4}{9} \, \text{°C} \] ### Conclusion: Final Answer Thus, the rise in temperature of the substance when it hits the ground is: **Final Answer: B. \( \frac{4}{9} \, \text{°C} \)** ### Explanation of Other Options - **Option A: \( \frac{2}{9} \, \text{°C} \)**: This value is too low and does not account for the total potential energy converted into heat. - **Option C: \( \frac{9}{4} \, \text{°C} \)**: This value is too high and suggests an incorrect relationship between potential energy and heat energy. - **Option D: \( \frac{9}{2} \, \text{°C} \)**: This is also too high and does not reflect the correct conversion of energy. ### Revision Summary - The potential energy of a falling object is calculated using \( PE = mgh \). - The heat energy gained by the substance is given by \( Q = mc\Delta T \). - Setting the potential energy equal to the heat energy allows us to find the rise in temperature. - The specific heat capacity is crucial in determining how much the temperature will rise based on the energy converted. This thorough understanding of energy conversion and the relationships between potential energy and heat energy is essential for solving similar problems in physics.