A 38 kg block of lead is heated from -26 degrees Celsius to 180 degrees Celsius. How much heat energy does the block of lead absorb to increase its temperature this much?

The specific heat capacity of lead is 0.128 J/g°C. First, calculate the change in temperature (180°C - (-26°C) = 206°C). Then, calculate the heat energy using the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. So, Q = 38 kg * 0.128 J/g°C * 206°C. This equals approximately 991.36 kJ of heat energy absorbed.