A handful of lead shot is heated to 80°C and dumped into 250.0g of water at 20.0°C. The mixture comes to thermal equilibrium at 40.0°C.?

Let m g be the mass of the lead.

specific heat of lead = 0.16 J/g°C
specific heat of water = 4.184 J/g°C

Heat change = mass × (specific heat) × (temperature change)
Heat lost by the lead = m × 0.16 × (80 - 40.0) J
Heat gained by the water = 250.0 × 4.184 × (40.0 - 20.0) J

Assume there is no energy loss to or energy gain from the surroundings.
Heat lost by the lead = Heat gained by the water
m × 0.16 × (80 - 40.0) J = 250.0 × 4.184 × (40.0 - 20.0) J
m = 3270

Hence, mass of the lead = 3270 g