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A total of 54.0 Joules of heat are observed as 58.3g of lead is heated from 12.0°C to 42.0°C. From this data, what is the specific heat of lead?

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Final answer:

The specific heat of lead is 0.031 J/(g°C).

Step-by-step explanation:

The specific heat is a property of a substance that represents the amount of heat required to raise the temperature of a unit mass (in grams) of that substance by one degree Celsius or Kelvin. To calculate the specific heat of lead, we can use the formula:

Q = m * c * ΔT

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

In this case, we have:

Q = 54.0 J, m = 58.3 g, ΔT = 42.0°C - 12.0°C = 30.0°C.

Substituting these values into the formula, we get:

54.0 J = 58.3 g * c * 30.0°C

To calculate the specific heat, c, we divide both sides of the equation by (58.3 g * 30.0°C):

c = 54.0 J / (58.3 g * 30.0°C) = 0.031 J/(g°C)

Therefore, the specific heat of lead is 0.031 J/(g°C).

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