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  

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