Respuesta :
Answer:
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
The mass of the sample of lead is 89.7 g
From the question,
We are to determine the mass of the sample (lead)
Using the equation,
Q = mcΔT
Where Q is the quantity of heat
m is the mass of substance
c is the specific heat capacity of substance
and ΔT is the change in temperature
From the given information
Q = 265 J
ΔT = 22.9 °C
and
Specific heat capacity of lead = 0.129 J/g °C
Putting the parameters into the formula, we
265 = m × 0.129 × 22.9
265 = m × 2.9541
∴ m = 265 ÷ 2.9541
m = 89.7058 g
m ≅ 89.7 g
Hence, the mass of the sample of lead is 89.7 g
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