# Calculating Heat Absorbed by an Aluminum Block ## Step 1 Understand the Formula The amount of heat \( Q \) absorbed or released is given by: \[ Q = m \cdot c \cdot \Delta T \] Where: - \( m \) = mass (kg) - \( c \) = specific heat capacity (J/kg·K) - \( \Delta T \) = temperature change (K or °C) ## Step 2: Substitute Values and Calculate Given: - \( m = 5 \) kg - \( c = 900 \) J/kg·K - Initial temperature \( T_1 = 25^\circ \)C - Final temperature \( T_2 = 150^\circ \)C Calculate temperature change: \[ \Delta T = T_2 - T_1 = 150^\circ\text{C} - 25^\circ\text{C} = 125\ \text{K} \] Plug into the formula: \[ Q = 5 \times 900 \times 125 = 562,\!500 \ \text{J} \] Convert joules to kilojoules: \[ Q = \frac{562,\!500}{1,\!000} = 562.5\ \text{kJ} \] ## Example Calculation Suppose a 2 kg aluminum block is heated from 30°C to 80°C. - \( m = 2 \) kg - \( c = 900 \) J/kg·K - \( \Delta T = 80 - 30 = 50 \) K \[ Q = 2 \times 900 \times 50 = 90,\!000\ \text{J} = 90\ \text{kJ} \] ## Graph: Heat Absorbed vs Temperature  *Graph: Heat energy absorbed increases linearly with temperature for a 5 kg aluminum block.* ## Final Answer **The total amount of heat energy absorbed by the aluminum block is \(\boxed{562.5\ \text{kJ}}\).**