Question

How much energy would be required to heat 25.78 g of lead from 20 to 140 degrees Celsius?

Answer 1

394.16 Joules

Step-by-step explanation:

Q = mcΔT

lead, the specific heat capacity (c) is approximately 0.128 J/g°C.

Given:

m = 25.78 g

ΔT = 140°C - 20°C = 120°C

c = 0.128 J/g°C

Plug in these values into the formula:

Q = 25.78 g * 0.128 J/g°C * 120°C

Q = 394.1632 J