The temperature of a sample of lead increased by 22.9 °C when 265 J of heat was applied. What is the mass of the sample?

90.41 g

Explanation:

Data provided in the question:

Increase in temperature of lead sample, ΔT = 22.9 °C

Heat supplied to the sample, Q = 265 J

Now,

we know that

Q = mCΔT

here,

m = mass of the sample

C = Specific heat of lead = 0.128 J/g.°C

Thus,

265 = m * 0.128 * 22.9

or

265 = m * 2.9312

or

m = 90.41 g