A sample of silicon absorbs 14.9 J of heat, upon which the temperature of the sample increases from
21.5°C to 28.5°C. If the specific heat of silicon is 0.710 J g
-1
K
-1
, what is the mass (in grams) of the
sample?
To find the mass of the sample, we can use the formula:
Q = mc∆T
where:
Q = heat absorbed = 14.9 J
m = mass of the sample (what we're trying to find)
c = specific heat of silicon = 0.710 J/g·K
∆T = change in temperature = (final temperature - initial temperature) = (28.5°C - 21.5°C) = 7°C
Rearranging the formula, we have:
m = Q / (c∆T)
Substituting in the values we have:
m = 14.9 J / (0.710 J/g·K * 7°C)
Simplifying:
m = 14.9 J / (4.97 J/g)
m ≈ 2.99 g
Therefore, the mass of the sample is approximately 2.99 grams.
To find the mass of the sample, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the amount of heat transferred,
m is the mass of the sample,
c is the specific heat of the material, and
ΔT is the change in temperature.
In this case, the heat transferred is given as 14.9 J, the specific heat of silicon is 0.710 J g^(-1)K^(-1), and the change in temperature is 28.5°C - 21.5°C = 7°C.
Now we can rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)
Substituting the values:
m = 14.9 J / (0.710 J g^(-1)K^(-1) * 7°C)
Simplifying the equation:
m = 14.9 J / (4.97 J g^(-1)K^(-1))
m ≈ 2.99 g
Therefore, the mass of the sample is approximately 2.99 grams.